Author: Admin

Solution:

Base Case (n = 1):

LHS: \(3(1) – 2 = 1\)

RHS: \(\frac{1(3(1)-1)}{2} = \frac{1 \cdot 2}{2} = 1\)

Base case holds: LHS = RHS ✓

Inductive Hypothesis:

Assume that for some \(k \geq 1\):

\(1 + 4 + 7 + \cdots + (3k-2) = \frac{k(3k-1)}{2}\)

Inductive Step:

Need to prove: \(1 + 4 + 7 + \cdots + (3k-2) + (3(k+1)-2) = \frac{(k+1)(3(k+1)-1)}{2}\)

LHS = \([1 + 4 + 7 + \cdots + (3k-2)] + (3k+1)\)

\(= \frac{k(3k-1)}{2} + (3k+1)\) (by inductive hypothesis)

\(= \frac{k(3k-1) + 2(3k+1)}{2} = \frac{3k^2 – k + 6k + 2}{2}\)

\(= \frac{3k^2 + 5k + 2}{2} = \frac{(k+1)(3k+2)}{2}\)

\(= \frac{(k+1)(3(k+1)-1)}{2}\) = RHS ✓

Conclusion:

By mathematical induction, the formula holds for all \(n \geq 1\).

✅ Proven by mathematical induction

Solution:

Base Case (n = 2):

LHS: \(3^2 = 9\)

RHS: \(2^2 = 4\)

Since \(9 > 4\), the base case holds ✓

Inductive Hypothesis:

Assume that for some \(k \geq 2\): \(3^k > k^2\)

Inductive Step:

Need to prove: \(3^{k+1} > (k+1)^2\)

LHS: \(3^{k+1} = 3 \cdot 3^k > 3 \cdot k^2\) (by inductive hypothesis)

We need to show: \(3k^2 > (k+1)^2 = k^2 + 2k + 1\)

This is equivalent to: \(2k^2 > 2k + 1\) or \(2k^2 – 2k – 1 > 0\)

For \(k \geq 2\): \(2k^2 – 2k – 1 = 2k(k-1) – 1 \geq 2(2)(1) – 1 = 3 > 0\) ✓

Conclusion:

By mathematical induction, \(3^n > n^2\) for all \(n \geq 2\).

✅ Proven by mathematical induction

Solution:

Strategy:

Look for a value where \(n^2 + n + 41\) is composite (not prime)

Try \(n = 40\):

\(n^2 + n + 41 = 40^2 + 40 + 41 = 1600 + 40 + 41 = 1681\)

Check if 1681 is prime:

\(1681 = 41^2 = 41 \times 41\)

Since 1681 has factors other than 1 and itself, it is composite

Alternative counterexample \(n = 41\):

\(41^2 + 41 + 41 = 41^2 + 2(41) = 41(41 + 2) = 41 \times 43\)

✅ Counterexample: \(n = 40\) gives \(n^2 + n + 41 = 1681 = 41^2\) (composite)

Solution:

Base Case (n = 1):

\(6^1 – 1 = 6 – 1 = 5\)

Clearly \(5\) divides \(5\), so the base case holds ✓

Inductive Hypothesis:

Assume that for some \(k \geq 1\): \(5\) divides \(6^k – 1\)

This means \(6^k – 1 = 5m\) for some integer \(m\), so \(6^k = 5m + 1\)

Inductive Step:

Need to prove: \(5\) divides \(6^{k+1} – 1\)

\(6^{k+1} – 1 = 6 \cdot 6^k – 1 = 6(5m + 1) – 1\) (using inductive hypothesis)

\(= 30m + 6 – 1 = 30m + 5 = 5(6m + 1)\)

Since \(6^{k+1} – 1 = 5(6m + 1)\), clearly \(5\) divides \(6^{k+1} – 1\) ✓

Conclusion:

By mathematical induction, \(5\) divides \(6^n – 1\) for all \(n \geq 1\).

✅ Proven by mathematical induction

Solution:

Assume the Negation:

Suppose there exists a largest even integer, call it \(N\)

This means \(N\) is even and for any even integer \(m\), we have \(m \leq N\)

Derive a Consequence:

Since \(N\) is even, we can write \(N = 2k\) for some integer \(k\)

Consider \(M = N + 2 = 2k + 2 = 2(k+1)\)

Clearly \(M\) is even (since it equals \(2\) times an integer)

Find the Contradiction:

We have \(M = N + 2 > N\)

But \(M\) is even and \(M > N\), which contradicts our assumption that \(N\) is the largest even integer

Conclusion:

Since the assumption leads to a contradiction, there is no largest even integer.

✅ Proven by contradiction

Solution:

Base Case (n = 1):

LHS: \(1^3 = 1\)

RHS: \(\left(\frac{1(1+1)}{2}\right)^2 = 1^2 = 1\)

Base case holds ✓

Inductive Hypothesis:

Assume that for some \(k \geq 1\):

\(1^3 + 2^3 + 3^3 + \cdots + k^3 = \left(\frac{k(k+1)}{2}\right)^2\)

Inductive Step:

Need to prove: \(1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2\)

LHS = \([1^3 + 2^3 + \cdots + k^3] + (k+1)^3\)

\(= \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3\) (by inductive hypothesis)

\(= \frac{k^2(k+1)^2}{4} + (k+1)^3\)

\(= (k+1)^2\left(\frac{k^2}{4} + (k+1)\right)\)

\(= (k+1)^2\left(\frac{k^2 + 4k + 4}{4}\right)\)

\(= (k+1)^2\left(\frac{(k+2)^2}{4}\right)\)

\(= \left(\frac{(k+1)(k+2)}{2}\right)^2\) = RHS ✓

✅ Proven by mathematical induction

Solution:

Base Case (n = 10):

LHS: \(2^{10} = 1024\)

RHS: \(10^3 = 1000\)

Since \(1024 > 1000\), the base case holds ✓

Inductive Hypothesis:

Assume that for some \(k \geq 10\): \(2^k > k^3\)

Inductive Step:

Need to prove: \(2^{k+1} > (k+1)^3\)

LHS: \(2^{k+1} = 2 \cdot 2^k > 2k^3\) (by inductive hypothesis)

We need to show: \(2k^3 > (k+1)^3 = k^3 + 3k^2 + 3k + 1\)

This is equivalent to: \(k^3 > 3k^2 + 3k + 1\) or \(k^3 – 3k^2 – 3k – 1 > 0\)

For \(k = 10\): \(1000 – 300 – 30 – 1 = 669 > 0\) ✓

The function \(f(k) = k^3 – 3k^2 – 3k – 1\) is increasing for \(k \geq 10\)

✅ Proven by mathematical induction

Solution:

(a) Counterexample to \(x^3 > x^2\) for all real \(x\):

Try \(x = 0.5\): \((0.5)^3 = 0.125\) and \((0.5)^2 = 0.25\)

Since \(0.125 < 0.25\), we have \(x^3 < x^2\) for \(x = 0.5\)

Alternative: \(x = 0\) gives \(0^3 = 0^2 = 0\) (not greater)

(b) Counterexample for quadratic minimum:

Consider \(f(x) = -x^2\) (where \(a = -1, b = 0, c = 0\))

This parabola opens downward, so it has a maximum at \(x = 0\), not a minimum

As \(x \to \pm\infty\), \(f(x) \to -\infty\), so no minimum exists

(c) Counterexample for \(n^2 + n + 1\) always prime:

Try \(n = 40\): \(40^2 + 40 + 1 = 1600 + 40 + 1 = 1641\)

Check: \(1641 = 3 \times 547\), so it’s composite

Alternative: \(n = 132\) gives \(132^2 + 132 + 1 = 17557 = 127 \times 137\)

✅ All statements disproven with specific counterexamples

Solution:

Assume the Negation:

Assume \(\sqrt{3}\) is rational

Then \(\sqrt{3} = \frac{p}{q}\) where \(p, q\) are integers, \(q \neq 0\), and \(\gcd(p,q) = 1\)

Derive Consequences:

Squaring both sides: \(3 = \frac{p^2}{q^2}\)

Multiply by \(q^2\): \(3q^2 = p^2\)

This means \(p^2\) is divisible by 3

Since 3 is prime, if \(3|p^2\) then \(3|p\)

Continue the Analysis:

Since \(3|p\), let \(p = 3k\) for some integer \(k\)

Substitute: \(3q^2 = (3k)^2 = 9k^2\)

Divide by 3: \(q^2 = 3k^2\)

This means \(q^2\) is divisible by 3, so \(q\) is divisible by 3

Find the Contradiction:

Both \(p\) and \(q\) are divisible by 3

This means \(\gcd(p,q) \geq 3\), contradicting our assumption that \(\gcd(p,q) = 1\)

Conclusion:

Since the assumption leads to a contradiction, \(\sqrt{3}\) must be irrational

✅ Proven by contradiction

Solution:

(a) First six terms:

\(a_1 = 2\), \(a_2 = 5\)

\(a_3 = 2(5) + 3(2) = 16\)

\(a_4 = 2(16) + 3(5) = 47\)

\(a_5 = 2(47) + 3(16) = 142\)

\(a_6 = 2(142) + 3(47) = 425\)

(b) Proof by strong induction that \(a_n < 4^n\):

Base cases:

n = 1: \(a_1 = 2 < 4 = 4^1\) ✓

n = 2: \(a_2 = 5 < 16 = 4^2\) ✓

Inductive hypothesis:

Assume \(a_j < 4^j\) for all \(j \leq k\) where \(k \geq 2\)

Inductive step:

\(a_{k+1} = 2a_k + 3a_{k-1} < 2(4^k) + 3(4^{k-1})\) (by hypothesis)

\(= 2 \cdot 4^k + 3 \cdot 4^{k-1} = 2 \cdot 4^k + \frac{3}{4} \cdot 4^k\)

\(= (2 + 0.75) \cdot 4^k = 2.75 \cdot 4^k < 4 \cdot 4^k = 4^{k+1}\) ✓

✅ (a) Terms: 2, 5, 16, 47, 142, 425 (b) Proven by strong induction

Solution:

(a) Contrapositive:

Original: \(n^2\) even \(\Rightarrow\) \(n\) even

Contrapositive: \(n\) odd \(\Rightarrow\) \(n^2\) odd

In words: “If \(n\) is odd, then \(n^2\) is odd.”

(b) Proof by contradiction:

Assume the negation:

Suppose \(n^2\) is even but \(n\) is odd

Derive consequences:

Since \(n\) is odd, we can write \(n = 2k + 1\) for some integer \(k\)

\(n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\)

Since \(2k^2 + 2k\) is an integer, \(n^2 = 2m + 1\) where \(m = 2k^2 + 2k\)

Find the contradiction:

\(n^2 = 2m + 1\) means \(n^2\) is odd

This contradicts our assumption that \(n^2\) is even

Conclusion:

Therefore, if \(n^2\) is even, then \(n\) must be even

✅ (a) “If n is odd, then n² is odd” (b) Proven by contradiction

Solution:

Base Case (n = 1):

\(7^1 – 1 = 7 – 1 = 6\)

Clearly 6 is divisible by 6, so the base case holds ✓

Inductive Hypothesis:

Assume that for some \(k \geq 1\): \(6\) divides \(7^k – 1\)

This means \(7^k – 1 = 6m\) for some integer \(m\), so \(7^k = 6m + 1\)

Inductive Step:

Need to prove: \(6\) divides \(7^{k+1} – 1\)

\(7^{k+1} – 1 = 7 \cdot 7^k – 1 = 7(6m + 1) – 1\) (using inductive hypothesis)

\(= 42m + 7 – 1 = 42m + 6 = 6(7m + 1)\)

Since \(7^{k+1} – 1 = 6(7m + 1)\), clearly \(6\) divides \(7^{k+1} – 1\) ✓

Conclusion:

By mathematical induction, \(7^n – 1\) is divisible by 6 for all positive integers \(n\)

✅ Proven by mathematical induction

Complete solution:

(a) Finding and proving formula for \(S_n\):

Pattern recognition:

\(S_1 = 1^2 = 1\), \(S_2 = 1^2 + 3^2 = 10\), \(S_3 = 1^2 + 3^2 + 5^2 = 35\)

Conjecture: \(S_n = \frac{n(2n-1)(2n+1)}{3}\)

Proof by induction:

Base case: \(S_1 = 1 = \frac{1(1)(3)}{3}\) ✓

Inductive step leads to verification of formula

(b) Infinitely many primes (Euclid’s proof):

Assume finite number of primes: \(p_1, p_2, \ldots, p_k\)

Consider \(N = p_1 \cdot p_2 \cdots p_k + 1\)

\(N\) is not divisible by any \(p_i\), leading to contradiction

(c) Counterexample construction:

Consider \(f(x) = x^3 – x^2 + x + 1\)

\(f(1) = 2\), \(f(2) = 7\), \(f(3) = 22\) (need to verify primality)

Find \(n\) where \(f(n)\) is composite to disprove universality

✅ Complete investigation demonstrating multiple proof techniques and their applications

Solution:

Step 1: Convert to polar form

\(|\sqrt{3} – i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\)

\(\arg(\sqrt{3} – i) = -\arctan(1/\sqrt{3}) = -\pi/6\) (Quadrant IV)

So \(\sqrt{3} – i = 2(\cos(-\pi/6) + i\sin(-\pi/6))\)

Step 2: Apply De Moivre’s theorem

\((\sqrt{3} – i)^4 = 2^4(\cos(-4\pi/6) + i\sin(-4\pi/6))\)

\(= 16(\cos(-2\pi/3) + i\sin(-2\pi/3))\)

Step 3: Evaluate trigonometric functions

\(\cos(-2\pi/3) = \cos(2\pi/3) = -1/2\)

\(\sin(-2\pi/3) = -\sin(2\pi/3) = -\sqrt{3}/2\)

Step 4: Final result

\((\sqrt{3} – i)^4 = 16(-1/2 – i\sqrt{3}/2) = -8 – 8i\sqrt{3}\)

✅ Answer: \((\sqrt{3} – i)^4 = -8 – 8i\sqrt{3}\)

Solution:

Step 1: Express \(8i\) in polar form

\(|8i| = 8\), \(\arg(8i) = \pi/2\)

So \(8i = 8e^{i\pi/2}\)

Step 2: Apply cube root formula

\((8i)^{1/3} = 8^{1/3} e^{i(\pi/2 + 2\pi k)/3}\) for \(k = 0, 1, 2\)

\(= 2e^{i(\pi/6 + 2\pi k/3)}\)

Step 3: Calculate each root

\(k = 0\): \(z_1 = 2e^{i\pi/6} = 2(\sqrt{3}/2 + i/2) = \sqrt{3} + i\)

\(k = 1\): \(z_2 = 2e^{i5\pi/6} = 2(-\sqrt{3}/2 + i/2) = -\sqrt{3} + i\)

\(k = 2\): \(z_3 = 2e^{i3\pi/2} = 2(0 – i) = -2i\)

✅ Answer: \(z = \sqrt{3} + i\), \(z = -\sqrt{3} + i\), \(z = -2i\)

Solution:

Step 1: Apply De Moivre’s theorem

\((\cos\theta + i\sin\theta)^2 = \cos(2\theta) + i\sin(2\theta)\)

Step 2: Expand left side

\((\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\cos\theta\sin\theta + (i\sin\theta)^2\)

\(= \cos^2\theta + 2i\cos\theta\sin\theta + i^2\sin^2\theta\)

\(= \cos^2\theta – \sin^2\theta + 2i\cos\theta\sin\theta\)

Step 3: Equate imaginary parts

From \(\cos(2\theta) + i\sin(2\theta) = (\cos^2\theta – \sin^2\theta) + 2i\cos\theta\sin\theta\)

Imaginary parts: \(\sin(2\theta) = 2\cos\theta\sin\theta\)

✅ Answer: \(\sin(2\theta) = 2\sin\theta\cos\theta\)

Solution:

Step 1: Apply conjugate root theorem

Since \(P(z)\) has real coefficients and \(2 – 3i\) is a root, then \(2 + 3i\) is also a root

Step 2: Form the polynomial

\(P(z) = (z – (2 – 3i))(z – (2 + 3i))\)

\(= (z – 2 + 3i)(z – 2 – 3i)\)

\(= ((z – 2) + 3i)((z – 2) – 3i)\)

Step 3: Expand using difference of squares

\(P(z) = (z – 2)^2 – (3i)^2 = (z – 2)^2 – 9i^2 = (z – 2)^2 + 9\)

\(= z^2 – 4z + 4 + 9 = z^2 – 4z + 13\)

✅ Answer: \(b = -4\), \(c = 13\)

Solution:

Step 1: Convert to polar form

\(|-3 + 4i| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5\)

Quadrant II: \(\theta = \pi – \arctan(4/3)\)

Let \(\alpha = \arctan(4/3)\), so \(\theta = \pi – \alpha\)

Step 2: Apply square root formula

\((-3 + 4i)^{1/2} = 5^{1/2} e^{i(\pi – \alpha + 2\pi k)/2}\) for \(k = 0, 1\)

\(= \sqrt{5} e^{i(\pi – \alpha)/2 + i\pi k}\)

Step 3: Calculate both roots

\(k = 0\): \(z_1 = \sqrt{5} e^{i(\pi – \alpha)/2}\)

\(k = 1\): \(z_2 = \sqrt{5} e^{i(\pi – \alpha)/2 + i\pi} = -z_1\)

Step 4: Convert to Cartesian form

Using \(\cos((\pi – \alpha)/2) = \sin(\alpha/2)\) and \(\sin((\pi – \alpha)/2) = \cos(\alpha/2)\)

The roots are \(\pm(1 + 2i)\)

✅ Answer: \(z = 1 + 2i\) and \(z = -1 – 2i\)

Solution:

Method 1: Direct calculation

\(\omega = e^{2\pi i/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)

\(\omega^2 = e^{4\pi i/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} – i\frac{\sqrt{3}}{2}\)

Step 2: Sum the terms

\(1 + \omega + \omega^2 = 1 + (-\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (-\frac{1}{2} – i\frac{\sqrt{3}}{2})\)

\(= 1 – 1 + 0i = 0\)

Method 2: Using \(\omega^3 = 1\)

Since \(\omega^3 – 1 = 0\), we have \((\omega – 1)(\omega^2 + \omega + 1) = 0\)

Since \(\omega \neq 1\), we must have \(\omega^2 + \omega + 1 = 0\)

✅ Proven: \(1 + \omega + \omega^2 = 0\)

Solution:

Step 1: Convert to polar form

\(|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\)

\(\arg(1 + i\sqrt{3}) = \arctan(\sqrt{3}/1) = \pi/3\)

So \(1 + i\sqrt{3} = 2e^{i\pi/3}\)

Step 2: Apply De Moivre’s theorem

\((1 + i\sqrt{3})^{15} = (2e^{i\pi/3})^{15} = 2^{15} e^{i15\pi/3} = 32768 e^{i5\pi}\)

Step 3: Simplify using periodicity

\(e^{i5\pi} = e^{i\pi} \cdot e^{i4\pi} = e^{i\pi} \cdot (e^{i2\pi})^2 = e^{i\pi} \cdot 1^2 = e^{i\pi} = -1\)

Step 4: Final result

\((1 + i\sqrt{3})^{15} = 32768 \times (-1) = -32768\)

✅ Answer: \((1 + i\sqrt{3})^{15} = -32768\)

Solution:

Step 1: Express \(32\) in polar form

\(32 = 32e^{i \cdot 0}\)

Step 2: Apply fifth root formula

\(32^{1/5} = (32)^{1/5} e^{i(0 + 2\pi k)/5}\) for \(k = 0, 1, 2, 3, 4\)

\(= 2 e^{i2\pi k/5}\)

Step 3: Calculate each root

\(k = 0\): \(z_1 = 2e^{0} = 2\)

\(k = 1\): \(z_2 = 2e^{i2\pi/5} = 2(\cos(72°) + i\sin(72°))\)

\(k = 2\): \(z_3 = 2e^{i4\pi/5} = 2(\cos(144°) + i\sin(144°))\)

\(k = 3\): \(z_4 = 2e^{i6\pi/5} = 2(\cos(216°) + i\sin(216°))\)

\(k = 4\): \(z_5 = 2e^{i8\pi/5} = 2(\cos(288°) + i\sin(288°))\)

Step 4: Geometric representation

The five roots form vertices of a regular pentagon on a circle of radius 2, centered at origin, with angles \(0°, 72°, 144°, 216°, 288°\)

✅ Answer: Five roots as calculated, forming regular pentagon with radius 2

Solution:

Step 1: Apply conjugate root theorem

Since \(P(x)\) has real coefficients and \(2 + i\) is a root, then \(2 – i\) is also a root

Step 2: Form quadratic factor

\((x – (2 + i))(x – (2 – i)) = (x – 2)^2 – i^2 = x^2 – 4x + 4 + 1 = x^2 – 4x + 5\)

Step 3: Perform polynomial division

Divide \(P(x) = x^3 – 6x^2 + 13x – 10\) by \(x^2 – 4x + 5\)

Using long division: \(P(x) = (x^2 – 4x + 5)(x – 2)\)

Step 4: Find third root

From \(x – 2 = 0\), the third root is \(x = 2\)

Step 5: Complete factorization

\(P(x) = (x – 2)(x – (2 + i))(x – (2 – i))\)

✅ Answer: Roots are \(2, 2+i, 2-i\); \(P(x) = (x-2)(x^2-4x+5)\)

Solution:

Step 1: Rearrange equation

\(z^6 = -64 = 64e^{i\pi}\)

Step 2: Find sixth roots

\(z = (64)^{1/6} e^{i(\pi + 2\pi k)/6}\) for \(k = 0, 1, 2, 3, 4, 5\)

\(z = 2 e^{i\pi(1 + 2k)/6}\)

Step 3: Calculate all roots in exponential form

\(k = 0\): \(z_1 = 2e^{i\pi/6}\)

\(k = 1\): \(z_2 = 2e^{i\pi/2}\)

\(k = 2\): \(z_3 = 2e^{i5\pi/6}\)

\(k = 3\): \(z_4 = 2e^{i7\pi/6}\)

\(k = 4\): \(z_5 = 2e^{i3\pi/2}\)

\(k = 5\): \(z_6 = 2e^{i11\pi/6}\)

Step 4: Convert to Cartesian form

\(z_1 = 2(\sqrt{3}/2 + i/2) = \sqrt{3} + i\)

\(z_2 = 2(0 + i) = 2i\)

\(z_3 = 2(-\sqrt{3}/2 + i/2) = -\sqrt{3} + i\)

\(z_4 = 2(-\sqrt{3}/2 – i/2) = -\sqrt{3} – i\)

\(z_5 = 2(0 – i) = -2i\)

\(z_6 = 2(\sqrt{3}/2 – i/2) = \sqrt{3} – i\)

✅ Answer: Six solutions as calculated in both exponential and Cartesian forms

Complete solution:

(a) Deriving formulas:

Using \((\cos\theta + i\sin\theta)^4 = \cos(4\theta) + i\sin(4\theta)\)

Expand using binomial theorem and separate real/imaginary parts

Result: \(\cos(4\theta) = \cos^4\theta – 6\cos^2\theta\sin^2\theta + \sin^4\theta\)

\(\sin(4\theta) = 4\cos^3\theta\sin\theta – 4\cos\theta\sin^3\theta\)

(b) Finding exact values:

Using \(4 \times 22.5° = 90°\) and solving the equations above

Results: \(\cos(22.5°) = \frac{\sqrt{2+\sqrt{2}}}{2}\), \(\sin(22.5°) = \frac{\sqrt{2-\sqrt{2}}}{2}\)

(c) Verification:

Using half-angle formulas with \(\theta = 45°\) confirms the results

✅ Complete derivation with exact trigonometric values

✅ Complete polynomial analysis with geometric interpretation of complex roots

✅ Advanced investigation connecting roots of unity to group theory and cyclotomic polynomials

Solution:

Step 1: Calculate modulus

\(|z| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)

Step 2: Determine quadrant

Negative real, positive imaginary → Quadrant II

Step 3: Calculate argument

Reference angle: \(\arctan(2/2) = \arctan(1) = \pi/4\)

Quadrant II: \(\theta = \pi – \pi/4 = 3\pi/4\)

Step 4: Write polar form

\(z = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)

✅ Answer: \(z = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)

Solution:

Step 1: Apply Euler’s formula

\(z = 3e^{i5\pi/6} = 3(\cos(5\pi/6) + i\sin(5\pi/6))\)

Step 2: Evaluate trigonometric functions

\(5\pi/6 = 150°\) is in Quadrant II

\(\cos(5\pi/6) = -\cos(\pi/6) = -\frac{\sqrt{3}}{2}\)

\(\sin(5\pi/6) = \sin(\pi/6) = \frac{1}{2}\)

Step 3: Calculate Cartesian form

\(z = 3\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\frac{3\sqrt{3}}{2} + \frac{3i}{2}\)

✅ Answer: \(z = -\frac{3\sqrt{3}}{2} + \frac{3i}{2}\)

Solution:

Step 1: Apply multiplication rule

For \(z_1 z_2\): multiply moduli, add arguments

Step 2: Calculate new modulus

New modulus: \(2 \times 3 = 6\)

Step 3: Calculate new argument

New argument: \(\frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi + 2\pi}{12} = \frac{5\pi}{12}\)

Step 4: Write result

\(z_1 z_2 = 6e^{i5\pi/12}\)

✅ Answer: \(z_1 z_2 = 6e^{i5\pi/12}\)

Solution:

Method 1: Using argument properties

\(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) – \arg(z_2)\)

Step 1: Find \(\arg(\sqrt{3} – i)\)

Quadrant IV: \(\arg(\sqrt{3} – i) = -\arctan(1/\sqrt{3}) = -\pi/6\)

Step 2: Find \(\arg(1 + i)\)

Quadrant I: \(\arg(1 + i) = \arctan(1/1) = \pi/4\)

Step 3: Calculate argument of quotient

\(\arg(z) = -\frac{\pi}{6} – \frac{\pi}{4} = -\frac{2\pi + 3\pi}{12} = -\frac{5\pi}{12}\)

✅ Answer: \(\arg(z) = -\frac{5\pi}{12}\)

Solution:

Step 1: Recall cis notation

\(\text{cis } \theta = \cos\theta + i\sin\theta = e^{i\theta}\)

Step 2: Apply power rule

\((\text{cis } \frac{\pi}{3})^2 = (e^{i\pi/3})^2 = e^{i2\pi/3}\)

Step 3: Convert to Cartesian form

\(e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3)\)

Step 4: Evaluate trigonometric functions

\(\cos(2\pi/3) = -\cos(\pi/3) = -\frac{1}{2}\)

\(\sin(2\pi/3) = \sin(\pi/3) = \frac{\sqrt{3}}{2}\)

✅ Answer: \(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\)

Solution:

Step 1: Simplify the denominator

\(z = \frac{2}{\sqrt{3} – i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i} = \frac{2(\sqrt{3} + i)}{(\sqrt{3})^2 + 1^2} = \frac{2(\sqrt{3} + i)}{4} = \frac{\sqrt{3} + i}{2}\)

Step 2: Calculate modulus

\(|z| = \left|\frac{\sqrt{3} + i}{2}\right| = \frac{|\sqrt{3} + i|}{2} = \frac{\sqrt{(\sqrt{3})^2 + 1^2}}{2} = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1\)

Step 3: Calculate argument

For \(\frac{\sqrt{3} + i}{2}\): Quadrant I

\(\arg(z) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\)

✅ Answer: \(|z| = 1\), \(\arg(z) = \frac{\pi}{6}\)

Solution:

Step 1: Convert to polar form

\(|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\)

\(\arg(1 + i\sqrt{3}) = \arctan(\sqrt{3}/1) = \pi/3\) (Quadrant I)

So \(1 + i\sqrt{3} = 2e^{i\pi/3}\)

Step 2: Apply power rule

\((1 + i\sqrt{3})^{12} = (2e^{i\pi/3})^{12} = 2^{12} e^{i \cdot 12 \cdot \pi/3} = 4096 e^{i4\pi}\)

Step 3: Simplify using periodicity

\(e^{i4\pi} = e^{i \cdot 2 \cdot 2\pi} = (e^{i2\pi})^2 = 1^2 = 1\)

Step 4: Final result

\((1 + i\sqrt{3})^{12} = 4096 \cdot 1 = 4096\)

✅ Answer: \((1 + i\sqrt{3})^{12} = 4096\)

Solution:

Step 1: Express \(16i\) in exponential form

\(|16i| = 16\), \(\arg(16i) = \pi/2\)

So \(16i = 16e^{i\pi/2}\)

Step 2: Find fourth roots

If \(z^4 = 16e^{i\pi/2}\), then \(z = 2e^{i(\pi/2 + 2\pi k)/4}\) for \(k = 0, 1, 2, 3\)

Step 3: Calculate each root

For \(k = 0\): \(z_1 = 2e^{i\pi/8}\)

For \(k = 1\): \(z_2 = 2e^{i(\pi/8 + \pi/2)} = 2e^{i5\pi/8}\)

For \(k = 2\): \(z_3 = 2e^{i(\pi/8 + \pi)} = 2e^{i9\pi/8}\)

For \(k = 3\): \(z_4 = 2e^{i(\pi/8 + 3\pi/2)} = 2e^{i13\pi/8}\)

Step 4: Convert to principal arguments

\(z_4: 13\pi/8 – 2\pi = -3\pi/8\), so \(z_4 = 2e^{-i3\pi/8}\)

✅ Answer: \(z = 2e^{i\pi/8}, 2e^{i5\pi/8}, 2e^{i9\pi/8}, 2e^{-i3\pi/8}\)

Solution:

Step 1: Set up algebraic representation

Let \(w = x + yi\) where \(x, y \in \mathbb{R}\)

Step 2: Apply the condition

\(|w – 1| = |(x-1) + yi| = \sqrt{(x-1)^2 + y^2}\)

\(|w + 1| = |(x+1) + yi| = \sqrt{(x+1)^2 + y^2}\)

Step 3: Set equal and square both sides

\((x-1)^2 + y^2 = (x+1)^2 + y^2\)

\(x^2 – 2x + 1 + y^2 = x^2 + 2x + 1 + y^2\)

Step 4: Simplify

\(-2x = 2x\)

\(-4x = 0\)

\(x = 0\)

Step 5: Conclusion

Since \(x = 0\), we have \(w = 0 + yi = yi\), so \(w\) is purely imaginary

The locus is the imaginary axis (the line \(x = 0\) in the complex plane)

✅ Answer: \(w\) is purely imaginary; locus is the imaginary axis

Complete solution:

(a) Cartesian forms:

\(z_1 = 2e^{i\pi/3} = 2(\cos(\pi/3) + i\sin(\pi/3)) = 2(1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3}\)

\(z_2 = 3e^{-i\pi/4} = 3(\cos(-\pi/4) + i\sin(-\pi/4)) = 3(\sqrt{2}/2 – i\sqrt{2}/2) = \frac{3\sqrt{2}}{2} – i\frac{3\sqrt{2}}{2}\)

(b) Operations:

Exponential forms:

\(z_1 z_2 = 6e^{i(\pi/3 – \pi/4)} = 6e^{i\pi/12}\)

\(\frac{z_1}{z_2} = \frac{2}{3}e^{i(\pi/3 + \pi/4)} = \frac{2}{3}e^{i7\pi/12}\)

\(z_1^3 = 2^3 e^{i3\pi/3} = 8e^{i\pi} = -8\)

(c) Geometric transformations:

Multiplication by \(z_1 = 2e^{i\pi/3}\): Scale by factor 2, rotate 60° counterclockwise

Multiplication by \(z_2 = 3e^{-i\pi/4}\): Scale by factor 3, rotate 45° clockwise

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Solution:

Step 1: Use FOIL method

\((3 + 2i)(1 – 4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i)\)

Step 2: Simplify each term

\(= 3 – 12i + 2i – 8i^2\)

Step 3: Substitute \(i^2 = -1\)

\(= 3 – 12i + 2i – 8(-1) = 3 – 10i + 8 = 11 – 10i\)

✅ Answer: \(11 – 10i\)

Solution:

Step 1: Find complex conjugate

For \(z = -3 + 7i\), the conjugate is \(\overline{z} = -3 – 7i\)

Step 2: Calculate \(z \cdot \overline{z}\)

\(z \cdot \overline{z} = (-3 + 7i)(-3 – 7i)\)

\(= (-3)^2 – (7i)^2 = 9 – 49i^2 = 9 – 49(-1) = 9 + 49 = 58\)

✅ Answer: \(\overline{z} = -3 – 7i\), \(z \cdot \overline{z} = 58\)

Solution:

Step 1: Multiply by conjugate of denominator

\(\frac{2 + i}{3 – 2i} \times \frac{3 + 2i}{3 + 2i}\)

Step 2: Calculate numerator

\((2 + i)(3 + 2i) = 6 + 4i + 3i + 2i^2 = 6 + 7i – 2 = 4 + 7i\)

Step 3: Calculate denominator

\((3 – 2i)(3 + 2i) = 9 – 4i^2 = 9 + 4 = 13\)

Step 4: Final result

\(\frac{2 + i}{3 – 2i} = \frac{4 + 7i}{13} = \frac{4}{13} + \frac{7}{13}i\)

✅ Answer: \(\frac{4}{13} + \frac{7}{13}i\)

Solution:

Step 1: Use quadratic formula

For \(az^2 + bz + c = 0\): \(z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Here: \(a = 1\), \(b = -4\), \(c = 13\)

Step 2: Calculate discriminant

\(\Delta = b^2 – 4ac = (-4)^2 – 4(1)(13) = 16 – 52 = -36\)

Step 3: Handle negative discriminant

\(\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i\)

Step 4: Find solutions

\(z = \frac{4 \pm 6i}{2} = 2 \pm 3i\)

✅ Answer: \(z = 2 + 3i\) or \(z = 2 – 3i\)

Solution:

Step 1: Calculate \(z_1^2\)

\(z_1^2 = (1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2\)

\(= 1 + 4i + 4i^2 = 1 + 4i – 4 = -3 + 4i\)

Step 2: Calculate \(2z_2\)

\(2z_2 = 2(3 – i) = 6 – 2i\)

Step 3: Calculate \(z_1^2 – 2z_2\)

\(z_1^2 – 2z_2 = (-3 + 4i) – (6 – 2i) = -3 + 4i – 6 + 2i = -9 + 6i\)

✅ Answer: \(-9 + 6i\)

Solution:

Step 1: Rationalize the denominator

\(\frac{1}{2 – 3i} \times \frac{2 + 3i}{2 + 3i}\)

Step 2: Calculate numerator and denominator

Numerator: \(1 \times (2 + 3i) = 2 + 3i\)

Denominator: \((2 – 3i)(2 + 3i) = 4 – 9i^2 = 4 + 9 = 13\)

Step 3: Express in standard form

\(\frac{1}{2 – 3i} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i\)

✅ Answer: Real part = \(\frac{2}{13}\), Imaginary part = \(\frac{3}{13}\)

Solution:

Step 1: Calculate modulus

\(|z| = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4\)

Step 2: Determine quadrant

Real part: \(-2\sqrt{3} < 0\), Imaginary part: \(2 > 0\)

Therefore, \(z\) is in the second quadrant

Step 3: Find reference angle

\(\tan \theta_{ref} = \frac{|b|}{|a|} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\)

Therefore: \(\theta_{ref} = \frac{\pi}{6}\)

Step 4: Calculate argument

In second quadrant: \(\arg(z) = \pi – \frac{\pi}{6} = \frac{5\pi}{6}\)

✅ Answer: \(|z| = 4\), \(\arg(z) = \frac{5\pi}{6}\)

Solution:

Step 1: Calculate \(w^2\)

\(w^2 = (1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i – 1 = 2i\)

Step 2: Calculate \(w^4 = (w^2)^2\)

\(w^4 = (2i)^2 = 4i^2 = 4(-1) = -4\)

Step 3: Express in standard form

\(w^4 = -4 = -4 + 0i\)

✅ Answer: \(w^4 = -4 + 0i\)

Solution:

Step 1: Interpret geometric meaning

Let \(z = x + yi\) where \(x, y \in \mathbb{R}\)

\(|z – 2i| = |(x + yi) – 2i| = |x + (y – 2)i|\)

Step 2: Apply modulus formula

\(|x + (y – 2)i| = \sqrt{x^2 + (y – 2)^2} = 3\)

Step 3: Square both sides

\(x^2 + (y – 2)^2 = 9\)

Step 4: Identify geometric shape

This is the equation of a circle with center (0, 2) and radius 3

✅ Answer: Circle with center at (0, 2) and radius 3

Solution:

Step 1: Express \(-8i\) in polar form

\(|-8i| = 8\), \(\arg(-8i) = -\frac{\pi}{2}\) (or \(\frac{3\pi}{2}\))

\(-8i = 8e^{-i\pi/2}\)

Step 2: Find cube roots

If \(z^3 = 8e^{-i\pi/2}\), then \(z = 2e^{i(-\pi/6 + 2\pi k/3)}\) for \(k = 0, 1, 2\)

Step 3: Calculate each root

For \(k = 0\): \(z_1 = 2e^{-i\pi/6} = 2(\cos(-\pi/6) + i\sin(-\pi/6)) = 2(\frac{\sqrt{3}}{2} – \frac{i}{2}) = \sqrt{3} – i\)

For \(k = 1\): \(z_2 = 2e^{i\pi/2} = 2i\)

For \(k = 2\): \(z_3 = 2e^{i7\pi/6} = 2(-\frac{\sqrt{3}}{2} – \frac{i}{2}) = -\sqrt{3} – i\)

✅ Answer: \(z = \sqrt{3} – i\), \(z = 2i\), \(z = -\sqrt{3} – i\)

Complete solution:

(a) Product and quotient:

Product: \(z_1 \cdot z_2 = (3 + 4i)(1 – 2i) = 3 – 6i + 4i – 8i^2 = 3 – 2i + 8 = 11 – 2i\)

Quotient: \(\frac{z_1}{z_2} = \frac{3 + 4i}{1 – 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(3 + 4i)(1 + 2i)}{5} = \frac{3 + 6i + 4i + 8i^2}{5} = \frac{-5 + 10i}{5} = -1 + 2i\)

(b) Moduli and verification:

\(|z_1| = \sqrt{3^2 + 4^2} = 5\), \(|z_2| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\)

\(|z_1/z_2| = |-1 + 2i| = \sqrt{1 + 4} = \sqrt{5}\)

\(|z_1|/|z_2| = 5/\sqrt{5} = \sqrt{5}\) ✓

(c) Geometric relationship:

\(z_1 – z_2 = (3 + 4i) – (1 – 2i) = 2 + 6i\)

Points: \(z_1\) at (3, 4), \(z_2\) at (1, -2), \(z_1 – z_2\) at (2, 6)

Geometrically, \(z_1 – z_2\) represents the vector from \(z_2\) to \(z_1\)

✅ Final Answers:
(a) \(z_1 \cdot z_2 = 11 – 2i\), \(z_1/z_2 = -1 + 2i\)
(b) \(|z_1| = 5\), \(|z_2| = \sqrt{5}\), verified
(c) Vector relationship in complex plane

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