1. (a) **Problem:** Express $1 + i\sqrt{3}$ and $1 - i\sqrt{3}$ in the form $re^{i\theta}$ where $-\pi < \theta \leq \pi$. Then evaluate $$\left(\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}\right)^{20}$$ in the form $a + ib$. **Step 1:** Find modulus and argument of $1 + i\sqrt{3}$. - Modulus: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. - Argument: $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$. So, $1 + i\sqrt{3} = 2e^{i\pi/3}$. **Step 2:** Find modulus and argument of $1 - i\sqrt{3}$. - Modulus: same as above, $2$. - Argument: $\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$. So, $1 - i\sqrt{3} = 2e^{-i\pi/3}$. **Step 3:** Compute the quotient: $$\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} = \frac{2e^{i\pi/3}}{2e^{-i\pi/3}} = e^{i\pi/3 - (-i\pi/3)} = e^{i2\pi/3}.$$ **Step 4:** Raise to the 20th power: $$\left(e^{i2\pi/3}\right)^{20} = e^{i40\pi/3} = e^{i(12\pi + 4\pi/3)} = e^{i4\pi/3}$$ (since $e^{i12\pi} = 1$). **Step 5:** Express $e^{i4\pi/3}$ in $a + ib$ form: $$a = \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \quad b = \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}.$$ **Answer:** $$\left(\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}\right)^{20} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}.$$ --- 1. (b) **Problem:** Given $(1 + i)^n = x + iy$, prove that $x^2 + y^2 = 2^n$. **Step 1:** Recall modulus of $1 + i$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$. **Step 2:** The modulus of $(1 + i)^n$ is $|1 + i|^n = (\sqrt{2})^n = 2^{n/2}$. **Step 3:** Since $x + iy = (1 + i)^n$, its modulus is $\sqrt{x^2 + y^2}$. **Step 4:** Equate moduli: $$\sqrt{x^2 + y^2} = 2^{n/2} \implies x^2 + y^2 = 2^n.$$ **Answer:** $x^2 + y^2 = 2^n$. --- 1. (c) **Problem:** Given $z = 4\sqrt{3}e^{i\pi/3} - 4e^{i5\pi/6}$, express $z$ in the form $re^{i\theta}$ and show $$\frac{z}{8} + i\left(\frac{z}{8}\right)^2 + \left(\frac{z}{8}\right)^3 = 2e^{i\pi/2}.$$ **Step 1:** Write $z$ in rectangular form: - $4\sqrt{3}e^{i\pi/3} = 4\sqrt{3}(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 4\sqrt{3}\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} + 6i$. - $4e^{i5\pi/6} = 4(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}) = 4\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -2\sqrt{3} + 2i$. **Step 2:** Compute $z$: $$z = (2\sqrt{3} + 6i) - (-2\sqrt{3} + 2i) = 2\sqrt{3} + 6i + 2\sqrt{3} - 2i = 4\sqrt{3} + 4i.$$ **Step 3:** Find modulus and argument of $z$: - Modulus: $r = \sqrt{(4\sqrt{3})^2 + 4^2} = \sqrt{48 + 16} = \sqrt{64} = 8$. - Argument: $\theta = \tan^{-1}\left(\frac{4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$. So, $z = 8e^{i\pi/6}$. **Step 4:** Compute the expression: $$\frac{z}{8} + i\left(\frac{z}{8}\right)^2 + \left(\frac{z}{8}\right)^3 = e^{i\pi/6} + i e^{i\pi/3} + e^{i\pi/2}.$$ **Step 5:** Write each term in rectangular form: - $e^{i\pi/6} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$. - $i e^{i\pi/3} = i(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = i\frac{1}{2} - \sin\frac{\pi}{3} = i\frac{1}{2} - i^2\frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$ (correcting: $i e^{i\pi/3} = i(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = i\frac{1}{2} + i^2 \frac{\sqrt{3}}{2} = i\frac{1}{2} - \frac{\sqrt{3}}{2}$). - $e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i1 = i$. **Step 6:** Sum real parts: $$\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 0 = 0.$$ Sum imaginary parts: $$\frac{1}{2} + \frac{1}{2} + 1 = 2.$$ **Step 7:** So the sum is $0 + 2i = 2i = 2e^{i\pi/2}$. **Answer:** Verified that $$\frac{z}{8} + i\left(\frac{z}{8}\right)^2 + \left(\frac{z}{8}\right)^3 = 2e^{i\pi/2}.$$ --- 2. (a) **Problem:** Find modulus and argument of $$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}$$ and show $$\left(\frac{1 + \sin\frac{\pi}{3} + i\cos\frac{\pi}{3}}{1 + \sin\frac{\pi}{3} - i\cos\frac{\pi}{3}}\right)^3 = i.$$ **Step 1:** Let numerator $N = 1 + \sin\theta + i\cos\theta$, denominator $D = 1 + \sin\theta - i\cos\theta$. **Step 2:** Compute modulus: - $|N| = \sqrt{(1 + \sin\theta)^2 + \cos^2\theta} = \sqrt{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta} = \sqrt{2 + 2\sin\theta}$. - $|D| = \sqrt{(1 + \sin\theta)^2 + (-\cos\theta)^2} = \sqrt{2 + 2\sin\theta}$. So modulus of fraction is 1. **Step 3:** Find argument: $$\arg\left(\frac{N}{D}\right) = \arg(N) - \arg(D).$$ - $\arg(N) = \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)$. - $\arg(D) = \tan^{-1}\left(\frac{-\cos\theta}{1 + \sin\theta}\right) = -\tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)$. So, $$\arg\left(\frac{N}{D}\right) = 2 \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right).$$ **Step 4:** Use identity: $$\tan\frac{\phi}{2} = \frac{\sin\phi}{1 + \cos\phi}.$$ Here, set $\phi = \frac{\pi}{2} - \theta$, then $$\tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right) = \frac{\pi}{4} - \frac{\theta}{2}.$$ Therefore, $$\arg\left(\frac{N}{D}\right) = 2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{\pi}{2} - \theta.$$ **Step 5:** For $\theta = \frac{\pi}{3}$, $$\arg = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}.$$ **Step 6:** Raise to power 3: $$\left(e^{i\pi/6}\right)^3 = e^{i\pi/2} = i.$$ **Answer:** Modulus is 1, argument is $\frac{\pi}{2} - \theta$, and the given expression cubed equals $i$. --- 2. (b) (i) **Problem:** Find modulus and argument of $$\left(\frac{2 + 3i}{3 + 2i}\right)^2.$$ **Step 1:** Compute modulus of numerator: $$|2 + 3i| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.$$ Denominator: $$|3 + 2i| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}.$$ **Step 2:** Modulus of fraction: $$\frac{\sqrt{13}}{\sqrt{13}} = 1.$$ **Step 3:** Argument of numerator: $$\tan^{-1}\left(\frac{3}{2}\right).$$ Argument of denominator: $$\tan^{-1}\left(\frac{2}{3}\right).$$ **Step 4:** Argument of fraction: $$\tan^{-1}\left(\frac{3}{2}\right) - \tan^{-1}\left(\frac{2}{3}\right).$$ **Step 5:** Square the fraction multiplies argument by 2: $$2\left(\tan^{-1}\left(\frac{3}{2}\right) - \tan^{-1}\left(\frac{2}{3}\right)\right).$$ **Answer:** Modulus is 1, argument is $2\left(\tan^{-1}(3/2) - \tan^{-1}(2/3)\right)$. --- 2. (b) (ii) **Problem:** Find modulus and argument of $$\sqrt{\frac{2 - i}{2 + i}}.$$ **Step 1:** Compute modulus of numerator: $$|2 - i| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}.$$ Denominator: $$|2 + i| = \sqrt{2^2 + 1^2} = \sqrt{5}.$$ **Step 2:** Modulus of fraction: $$\frac{\sqrt{5}}{\sqrt{5}} = 1.$$ **Step 3:** Argument of numerator: $$\tan^{-1}\left(\frac{-1}{2}\right) = -\tan^{-1}\left(\frac{1}{2}\right).$$ Argument of denominator: $$\tan^{-1}\left(\frac{1}{2}\right).$$ **Step 4:** Argument of fraction: $$-\tan^{-1}\left(\frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right) = -2\tan^{-1}\left(\frac{1}{2}\right).$$ **Step 5:** Square root halves the argument: $$-\tan^{-1}\left(\frac{1}{2}\right).$$ **Answer:** Modulus is 1, argument is $-\tan^{-1}(1/2)$. --- 2. (c) (i) **Problem:** Express $$\left(\frac{2 - i}{2 + i}\right)^2$$ in the form $a + bi$. **Step 1:** Compute numerator and denominator: $$\frac{2 - i}{2 + i} = \frac{(2 - i)(2 - i)}{(2 + i)(2 - i)} = \frac{(2 - i)^2}{2^2 + 1^2} = \frac{(2 - i)^2}{5}.$$ **Step 2:** Expand numerator: $$(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i.$$ **Step 3:** So fraction is $$\frac{3 - 4i}{5} = \frac{3}{5} - \frac{4}{5}i.$$ **Step 4:** Square the fraction: $$\left(\frac{3}{5} - \frac{4}{5}i\right)^2 = \left(\frac{3}{5}\right)^2 - 2\cdot\frac{3}{5}\cdot\frac{4}{5}i + \left(-\frac{4}{5}i\right)^2 = \frac{9}{25} - \frac{24}{25}i - \frac{16}{25}.$$ **Step 5:** Simplify real part: $$\frac{9}{25} - \frac{16}{25} = -\frac{7}{25}.$$ **Answer:** $$-\frac{7}{25} - \frac{24}{25}i.$$ --- 2. (c) (ii) **Problem:** Express $$\frac{(1 + 5i)(2 - 3i)}{(3 - i)(2 + 3i)}$$ in the form $a + bi$. **Step 1:** Compute numerator: $$(1 + 5i)(2 - 3i) = 2 - 3i + 10i - 15i^2 = 2 + 7i + 15 = 17 + 7i.$$ **Step 2:** Compute denominator: $$(3 - i)(2 + 3i) = 6 + 9i - 2i - 3i^2 = 6 + 7i + 3 = 9 + 7i.$$ **Step 3:** Write fraction: $$\frac{17 + 7i}{9 + 7i}.$$ **Step 4:** Multiply numerator and denominator by conjugate of denominator: $$\frac{(17 + 7i)(9 - 7i)}{(9 + 7i)(9 - 7i)} = \frac{153 - 119i + 63i - 49i^2}{81 + 49} = \frac{153 - 56i + 49}{130} = \frac{202 - 56i}{130}.$$ **Step 5:** Simplify: $$\frac{202}{130} - \frac{56}{130}i = \frac{101}{65} - \frac{28}{65}i.$$ **Answer:** $$\frac{101}{65} - \frac{28}{65}i.$$ --- 3. (a) **Problem:** Find all roots of $$z^6 + 6z^4 + 11z^2 + 6 = 0.$$ **Step 1:** Substitute $w = z^2$, then $$w^3 + 6w^2 + 11w + 6 = 0.$$ **Step 2:** Factor cubic polynomial: Try $w = -1$: $$(-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0.$$ So, $(w + 1)$ is a factor. **Step 3:** Divide polynomial by $(w + 1)$: $$w^3 + 6w^2 + 11w + 6 = (w + 1)(w^2 + 5w + 6).$$ **Step 4:** Factor quadratic: $$w^2 + 5w + 6 = (w + 2)(w + 3).$$ **Step 5:** Roots for $w$ are $-1, -2, -3$. **Step 6:** Recall $w = z^2$, so $$z^2 = -1, \quad z^2 = -2, \quad z^2 = -3.$$ **Step 7:** Find roots for each: - $z^2 = -1 \implies z = \pm i$. - $z^2 = -2 \implies z = \pm i\sqrt{2}$. - $z^2 = -3 \implies z = \pm i\sqrt{3}$. **Answer:** Roots are $$\pm i, \pm i\sqrt{2}, \pm i\sqrt{3}.$$ --- 3. (b) **Problem:** For $$(1 + i)z^2 - 2iz + 3 + i = 0,$$ find (i) $\alpha + \beta$ and $\alpha\beta$ in form $p + qi$, (ii) quadratic with roots $\alpha + 3\beta$ and $3\alpha + \beta$. **Step 1:** Coefficients: $$a = 1 + i, \quad b = -2i, \quad c = 3 + i.$$ **Step 2:** Sum of roots: $$\alpha + \beta = -\frac{b}{a} = -\frac{-2i}{1 + i} = \frac{2i}{1 + i}.$$ Multiply numerator and denominator by conjugate: $$\frac{2i(1 - i)}{(1 + i)(1 - i)} = \frac{2i - 2i^2}{1 + 1} = \frac{2i + 2}{2} = 1 + i.$$ **Step 3:** Product of roots: $$\alpha\beta = \frac{c}{a} = \frac{3 + i}{1 + i}.$$ Multiply numerator and denominator by conjugate: $$\frac{(3 + i)(1 - i)}{(1 + i)(1 - i)} = \frac{3 - 3i + i - i^2}{2} = \frac{3 - 2i + 1}{2} = \frac{4 - 2i}{2} = 2 - i.$$ **Answer (i):** $$\alpha + \beta = 1 + i, \quad \alpha\beta = 2 - i.$$ **Step 4:** Roots for new quadratic are $$r_1 = \alpha + 3\beta, \quad r_2 = 3\alpha + \beta.$$ **Step 5:** Sum: $$r_1 + r_2 = (\alpha + 3\beta) + (3\alpha + \beta) = 4(\alpha + \beta) = 4(1 + i) = 4 + 4i.$$ **Step 6:** Product: $$r_1 r_2 = (\alpha + 3\beta)(3\alpha + \beta) = 3\alpha^2 + \alpha\beta + 9\alpha\beta + 3\beta^2 = 3(\alpha^2 + \beta^2) + 10\alpha\beta.$$ **Step 7:** Use identity: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1 + i)^2 - 2(2 - i).$$ Calculate: $$(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i,$$ so $$\alpha^2 + \beta^2 = 2i - 2(2 - i) = 2i - 4 + 2i = -4 + 4i.$$ **Step 8:** Compute product: $$3(-4 + 4i) + 10(2 - i) = -12 + 12i + 20 - 10i = 8 + 2i.$$ **Step 9:** Quadratic with roots $r_1, r_2$ is $$z^2 - (r_1 + r_2)z + r_1 r_2 = 0,$$ which is $$z^2 - (4 + 4i)z + (8 + 2i) = 0.$$ **Answer (ii):** $$z^2 - (4 + 4i)z + (8 + 2i) = 0.$$ --- 3. (c) **Problem:** Given $$z_2^2 - z_1 z_2 + z_1^2 = 0,$$ with $z_1 = a + bi$, show $$z_2 = \frac{1}{2}(a - b\sqrt{3}) + \frac{1}{2}(b + a\sqrt{3})i$$ is a root. **Step 1:** Treat $z_2$ as unknown, quadratic in $z_2$: $$z_2^2 - z_1 z_2 + z_1^2 = 0.$$ **Step 2:** Use quadratic formula: $$z_2 = \frac{z_1 \pm \sqrt{z_1^2 - 4 z_1^2}}{2} = \frac{z_1 \pm \sqrt{-3 z_1^2}}{2} = \frac{z_1 \pm i \sqrt{3} z_1}{2} = \frac{z_1}{2}(1 \pm i\sqrt{3}).$$ **Step 3:** Substitute $z_1 = a + bi$: $$z_2 = \frac{a + bi}{2}(1 \pm i\sqrt{3}) = \frac{1}{2}[(a + bi)(1 \pm i\sqrt{3})].$$ **Step 4:** Expand for $+$ sign: $$(a + bi)(1 + i\sqrt{3}) = a + ai\sqrt{3} + bi + bi^2 \sqrt{3} = a + ai\sqrt{3} + bi - b\sqrt{3}$$ (since $i^2 = -1$). Group real and imaginary parts: - Real: $a - b\sqrt{3}$ - Imaginary: $a\sqrt{3} + b$ times $i$. **Step 5:** So $$z_2 = \frac{1}{2}(a - b\sqrt{3}) + \frac{1}{2}(b + a\sqrt{3})i,$$ which matches the given expression. **Answer:** The given $z_2$ is a root of the equation.