Question 11 Mark
Write the values of the square root of -i.
AnswerLet $\Rightarrow\sqrt-{\text{i}}=\text{x}+\text{iy}$ (Squaring both sides) $\Rightarrow-\text{i}=\text{x}^2+\text{y}^2\text{i}^2+2\text{ixy}$ $\Rightarrow2\text{xy}=-1 \ ...(\text{i})$ and $\text{x}^2-\text{y}^2=0 \ ...(\text{ii})$ By equation (ii), we find that x and y are of opposite sign. From equation (ii), $\text{x}=\pm\text{y}$ From equation (i), $2(\text{x})(\text{-x})=\frac{-1}{2}$ $\Rightarrow\text{x}=\pm\frac{1}{\sqrt2}$ $\Big[$ Since x and y have opposite signs, $\text{y}=-\frac{1}{\sqrt{2}}$ when $\text{x}=\frac{1}{\sqrt{2}}$ and vice versa $\Big]$ $\therefore \sqrt-{\text{i}}=\pm\frac{1}{\sqrt{2}}(1-\text{i})$
View full question & answer→Question 21 Mark
If $|\text{z}-5\text{i}|=|\text{z}+5\text{i}|,$ then find the locus of z.
Answer$|\text{z}-5\text{i}|=|\text{z}+5\text{i}|$ $\Rightarrow|\text{z}-5\text{i}|^2=|\text{z}+5\text{i}|^2$ $\Rightarrow(\text{z}-5\text{i})(\overline{\text{z}-5\text{i}})=(\text{z}+5\text{i})(\overline{\text{z}+5\text{i}}) \ \big[\because\text{z}\bar{\text{z}}=|\text{z}|^2\big]$ $\Rightarrow(\text{z}-5\text{i})(\bar{\text{z}}+5\text{i})=(\text{z}+5\text{i})(\bar{\text{z}}-5\text{i})$ $\Rightarrow\text{z}\bar{\text{z}}+5\text{zi}-5\bar{\text{z}}\text{i}-25\text{i}^2=\text{z}\bar{\text{z}}-5\text{zi}+5\bar{\text{z}}\text{i}-25\text{i}^2$ $\Rightarrow5\text{zi}+5\text{zi}=5\bar{\text{z}}\text{i}+5\bar{\text{z}}\text{i}$ $\Rightarrow10\text{zi}=10\bar{\text{z}}\text{i}$ $\Rightarrow\text{z}=\bar{\text{z}}$ $\Rightarrow\text{z}$ is purely real Hence, the locus of z is real axis.
View full question & answer→Question 31 Mark
Write $-1 + \text{i}\sqrt{3}$ in polar form.
AnswerLet $\text{z}=-1 + \text{i}\sqrt{3}.$ Then, $\text{r}=|\text{z}|=\sqrt{[-1]^2+[\sqrt{3}]^2}=2$ Let $\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$ $=\sqrt{3}$ $\Rightarrow\alpha=\frac{\pi}{3}$ Since the point representing z lies in the second quadrant. Therefore, the argument of z is given by $\theta=\pi-\alpha$ $=\pi-\frac{\pi}{3}$ $=\frac{2\pi}{3}$ So, the polar form is $\text{r}(\cos\theta+\text{i}\sin\theta)$ $\therefore\text{z}=2\Big(\cos\frac{2\pi}{3}+\text{i}\sin\frac{2\pi}{3}\Big)$
View full question & answer→Question 41 Mark
Write the least positive integral value of n for which $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{n}}$ is real.
Answer$\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{n}}$ $=\Big(\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}\Big)^{\text{n}}$ $=\Big(\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}\Big)^{\text{n}}$ $=\Big(\frac{1-1+2\text{i}}{1+1}\Big)^{\text{n}}$ $\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^{\text{n}}$ $=\text{i}^\text{n}$ For $\text{i}^\text{n}$ to be real, the smallest positive value of n will be 2. As, $\text{i}^2=-1,$ which is real.
View full question & answer→Question 51 Mark
Write the sum of the series $\text{i}+\text{i}^2+\text{i}^3+...$ upto 1000 terms.
AnswerWe know that $\text{i}+\text{i}^2+\text{i}^3+\text{i}^4=\text{i}-1-\text{i}+1=0$ $\therefore\text{i}+\text{i}^2+\text{i}^3+...+\text{i}^{1000}$ $=(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4)+(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8)...+\text{i}^{997}+\text{i}^{998}+\text{i}^{999}+\text{i}^{1000})$ $=(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4)+(\text{i}^4\text{i}+\text{i}^4\text{i}^2+\text{i}^4\text{i}^3+\text{i}^4\text{i}^4)...+\big[(\text{i}^4)^{249}\text{i}+(\text{i}^4)^{249}\text{i}^2+(\text{i}^4)^{249}\text{i}^3+(\text{i}^4)^{249}\text{i}^4\big]$ $=(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4)+(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4)+...+(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4)$ $=0$ Thus, the sum of the series $\text{i}+\text{i}^2+\text{i}^3+...$ upto 1000 terms is 0.
View full question & answer→Question 61 Mark
Write the argument of $(1+\text{i}\sqrt{3})(1+\text{i})(\cos\theta+\text{i}\sin\theta).$
AnswerLet the argument of $(1+\text{i}\sqrt{3})$ be $\alpha.$ Then, $\tan\alpha=\frac{\sqrt{3}}{1}=\tan\frac{\pi}{3}$ $\Rightarrow\alpha=\frac{\pi}{3}$ Let the argument of $(1+\text{i})$ be $\beta.$ Then, $\tan\beta=\frac{1}{1}=\tan\frac{\pi}{4}$ $\Rightarrow\beta=\frac{\pi}{3}$ Let the argument of $(\cos\theta+\text{i}\sin\theta)$ be $\gamma$ Then, $\tan\gamma=\frac{\sin\theta}{\cos\theta}=\tan\theta$ $\Rightarrow\gamma=\theta$ $\therefore$ The argument of $(1+\text{i}\sqrt{3})(1+\text{i})(\cos\theta+\text{i}\sin\theta)=\alpha+\beta+\gamma=\frac{\pi}{3}+\frac{\pi}{4}+\theta=\frac{7\pi}{12} +\theta$ Hence, the argument of $(1+\text{i}\sqrt{3})(1+\text{i})(\cos\theta+\text{i}\sin\theta)$ is $\frac{7\pi}{12}+\theta.$
View full question & answer→Question 71 Mark
Write the value of $\sqrt{-25}\times\sqrt{-9}.$
Answer$\sqrt{-25}\times\sqrt{-9}=5\sqrt{-1}\times3\sqrt{-1}$ $=5\text{i}\times3\text{i}$ $=15\text{i}^2$ $=-15$ Hence, $\sqrt{-25}\times\sqrt{-9}=-15.$
View full question & answer→Question 81 Mark
Write the value of $\text{arg(z)}+\text{arg}\bar{(\text{z})}.$
AnswerLet z be a complex number with argument $\theta.$ Then, $\text{z}=\text{re}^{\text{i}\theta}$ $\Rightarrow\bar{\text{z}}=\overline{\text{re}^{\text{i}\theta}}=\text{re}^{-\text{i}\theta}$ ⇒ argument of $\bar{\text{z}}$ is $\theta.$ Thus, $\text{arg(z)}+\text{arg}\bar{(\text{z})}=0.$
View full question & answer→Question 91 Mark
If n is any positive integer, write the value of $\frac{\text{i}^{4\text{n}+1}-\text{i}^{4\text{n}-1}}{2}.$
Answer$\frac{\text{i}^{4\text{n}+1}-\text{i}^{4\text{n}-1}}{2}$ $=\frac{\text{i}-\frac{1}{\text{i}}}{2} \ \big(\because\text{i}^{4\text{n}}=1,\text{i}^{-1}=\frac{1}{\text{i}}\big)$ $=\frac{\frac{\text{i}^2-1}{\text{i}}}{2}$ $=\frac{\text{i}^2-1}{2\text{i}}$ $=\frac{-1-1}{2\text{i}}$ $=\frac{-2}{-2\text{i}}$ $=\frac{-1}{\text{i}}$ $=\frac{-\text{i}}{\text{i}^2} \ \big(\because\text{i}^2=-1\big)$ $=\frac{-\text{i}}{-1}$ $=\text{i}$
View full question & answer→Question 101 Mark
Write the values of the square root of i.
AnswerLet the square root of i be $\text{x}+\text{iy}.$ $\Rightarrow\sqrt{\text{i}}=\text{x}+\text{iy}$ $\Rightarrow\text{i}=\text{x}^2+\text{y}^2\text{i}^2+2\text{ixy}$ $\Rightarrow\text{i}=\text{x}^2-\text{y}^2+2\text{ixy}$ (Squaring both sides) Comparing both the sides: $\text{x}^2-\text{y}^2=0 \ ...(\text{i})$ and $2\text{xy}=1 \ ...(\text{ii})$ By equation (ii), we find that x and y are of the same sign. From equation (i), $\text{x}=\pm\text{y}$ $\therefore\text{xy}=\frac{1}{2},\text{x}^2=\frac{1}{2}$ $\text{x}=\pm\frac{1}{2},\text{y}=\pm\frac{1}{\sqrt2}$ $\therefore \sqrt{\text{i}}=\pm\frac{1}{\sqrt{2}}(1+\text{i})$
View full question & answer→Question 111 Mark
Write $1 - \text{i}$ in polar form.
Answer$\text{z}=1 - \text{i}$ $\text{r}=|\text{z}|$ $=\sqrt{1+1}$ $=\sqrt{2}$ Let $\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$ $\therefore\tan\alpha=\Big|\frac{-1}{1}\Big|$ $=\frac{\pi}{4}$ $\Rightarrow\alpha=\frac{\pi}{4}$ Since point (1,−1) lies in the fourth quadrant, the argument of z is given by $\theta=-\alpha=-\frac{\pi}{4}$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$ $\sqrt{2}\Big\{\cos\big(-\frac{\pi}{4}\big)+\text{i}\sin\big(-\frac{\pi}{4}\big)\Big\}$ $\sqrt{2}\big(\cos\frac{\pi}{4}-\text{i}\sin\frac{\pi}{4}\big)$
View full question & answer→Question 121 Mark
If $\text{x}+\text{iy}=\sqrt{\frac{\text{a}+\text{ib}}{\text{c}+\text{id}}},$ then write the value of $(\text{x}^2+\text{y}^2)^2.$
Answer$\text{x}+\text{iy}=\sqrt{\frac{\text{a}+\text{ib}}{\text{c}+\text{id}}}$Taking modulus on both the sides,
$|\text{x}+\text{iy}|=\Big|\sqrt{\frac{\text{a}+\text{ib}}{\text{c}+\text{id}}}\Big|$
$\Rightarrow|\text{x}+\text{iy}|=\sqrt{\frac{|\text{a}+\text{ib}|}{|\text{c}+\text{id}|}}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{\frac{\sqrt{\text{a}^2+\text{b}^2}}{\sqrt{\text{c}^2+\text{d}^2}}} \ \Big[\because|\text{x}+\text{iy}|=\sqrt{\text{x}^2+\text{y}^2}\Big]$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{c}^2+\text{d}^2}}$
Squaring again, we get,
$\Rightarrow(\text{x}^2+\text{y}^2)^2=\frac{\text{a}^2+\text{b}^2}{\text{c}^2+\text{d}^2}$
View full question & answer→Question 131 Mark
If $\text{n}\in\text{N},$ then find the value of $\text{i}^\text{n}+\text{i}^{\text{n}+1}+\text{i}^{\text{n}+2}+\text{i}^{\text{n}+3}.$
Answer$\text{i}^\text{n}+\text{i}^{\text{n}+1}+\text{i}^{\text{n}+2}+\text{i}^{\text{n}+3}$ $=\text{i}^\text{n}+\text{i}^{\text{n}}.\text{i}+\text{i}^{\text{n}}.\text{i}^2+\text{i}^{\text{n}}.\text{i}^3$ $=\text{i}^\text{n}+\text{i}^{\text{n}}.\text{i}+\text{i}^{\text{n}}.(-1)+\text{i}^{\text{n}}.(-\text{i})$ $=\text{i}^\text{n}+\text{i}^{\text{n}}.\text{i}-\text{i}^{\text{n}}-\text{i}^{\text{n}}.\text{i}$ $=0$ Thus, the value of $\text{i}^\text{n}+\text{i}^{\text{n}+1}+\text{i}^{\text{n}+2}+\text{i}^{\text{n}+3}$ is 0.
View full question & answer→Question 141 Mark
Find the principal argument of $\Big(1+\text{i}\sqrt{3}\Big)^2.$
AnswerLet $\text{z}=\Big(1+\text{i}\sqrt{3}\Big)^2$ $=1+3\text{i}^2+2\sqrt{3}\text{i}$ $=1-3+2\sqrt{3}\text{i}$ $=-2+2\sqrt{3}\text{i}$ Let $\beta$ be an acute angle given by $\tan\beta=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|.$ Then, $\tan\beta=\Big|\frac{|2\sqrt{3}|}{|2|}\Big|=\big|\sqrt{3}\big|$ $\Rightarrow\tan\beta=\big|\tan\frac{\pi}{3}\big|$ $\Rightarrow\beta=\frac{\pi}{3}$ Clearly, z lies in the second quadrant. Therefore, the argument of z is given by $\text{arg(z)}=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ Hence, the principal argument of z is $\frac{2\pi}{3}.$
View full question & answer→Question 151 Mark
Write the argument of -i.
AnswerLet $\text{z}=-\text{i}$ Then, $\text{Re(z)}=0,\text{Im(z)}=-1$ Since, the point (0,-1) representing z = 0 -i lies on negative direction of imaginary axis.Therefore, $\text{arg(z)}=-\frac{\pi}{2}$ or $\frac{3\pi}{2}$
View full question & answer→Question 161 Mark
Express the following complex numbers in the standard form a + ib: $\frac{3+2\text{i}}{-2+\text{i}}$
Answer$\frac{3+2\text{i}}{-2+\text{i}}=\frac{3+2\text{i}}{(-2+\text{i})}\times\frac{(-2-\text{i})}{-2-\text{i}}$ [Rationalising the denominator] $=\frac{3(-2-\text{i})+2\text{i}(-2-\text{i})}{(-2)^2-(\text{i}^2)} \ \big[\because \ (\text{a}+\text{ib})(\text{a}-\text{ib})=\text{a}^2+\text{b}^2\big]$ $=\frac{-6-3\text{i}-4\text{i}+2}{4+1} \ \big[\because \ -\text{i}^2=1\big]$ $=\frac{-4-7\text{i}}{5}$ $=\frac{-4}{5}-\frac{7}{5}\text{i}$ $\therefore \ \frac{3+2\text{i}}{-2+\text{i}}=\frac{-4}{5}-\frac{7}{5}\text{i}$
View full question & answer→Question 171 Mark
Find the real value of a for which $3\text{i}^3-2\text{ai}^2+(1-\text{a})\text{i}+5$ is real.
Answer$3\text{i}^3-2\text{ai}^2+(1-\text{a})\text{i}+5$ $=-3\text{i}+2\text{a}+(1-\text{a})\text{i}+5$ $=(2\text{a}+5)+\text{i}(1-\text{a}-3)$ $=(2\text{a}+5)+\text{i}(-2-\text{a})$ Since, $3\text{i}^3-2\text{ai}^2+(1-\text{a})\text{i}+5$ is real. $\therefore\text{Im}[3\text{i}^3-2\text{ai}^2+(1-\text{a})\text{i}+5]=0$ $\Rightarrow-2-\text{a}=0$ $\Rightarrow\text{a}=-2$ Hence, the real value of a for which $3\text{i}^3-2\text{ai}^2+(1-\text{a})\text{i}+5$ is real is -2.
View full question & answer→Question 181 Mark
Write the value of $\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}.$
Answer$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}$ $=\frac{\text{i}^{4\times148}+\text{i}^{4\times147+2}+\text{i}^{4\times147}+\text{i}^{4\times146+2}+\text{i}^{4\times146}}{\text{i}^{4\times145+2}+\text{i}^{4\times145}+\text{i}^{4\times144+2}+\text{i}^{4\times144}+\text{i}^{4\times143+2}}$ $=\frac{1+\text{i}^2+1+\text{i}^2+1}{\text{i}^2+1+\text{i}^2+1+\text{i}^2} \ \big[\because\text{i}^4=1\big]$ $=\frac{1-1+1-1+1}{-1+1-1+1-1} \ \big[\because\text{i}^2=1\big]$ $=\frac{1}{-1}$ $=-1$
View full question & answer→Question 191 Mark
If $|\text{z}+4|\leq3,$ then find the greatest and least values of $|\text{z}+1|.$
Answer$|\text{z}+1|=|\text{z}+4-3|$ $\leq|\text{z}+4|+|-3|$ $\leq3+3$ $=6$ Also, $|\text{z}+1|\geq0$ Thus, $0\leq|\text{z}+1|\leq6.$ Hence, the greatest and least values of $|\text{z}+1|$ is 6 and 0.
View full question & answer→Question 201 Mark
For any two complex numbers $z_1$ and $z_2$ and any two real numbers a, b, find the value of $|\text{az}_1-\text{bz}_2|^2+|\text{az}_2+\text{bz}_1|^2.$
Answer$|\text{az}_1-\text{bz}_2|^2+|\text{az}_2+\text{bz}_1|^2\\=(\text{az}_1-\text{bz}_2)(\overline{\text{az}_1-\text{bz}_2})+(\text{az}_2+\text{bz}_1)(\overline{\text{az}_2+\text{bz}_1})$ $=(\text{az}_1-\text{bz}_2)(\text{a}\bar{\text{z}_1}-\text{b}\bar{\text{z}_2})+(\text{az}_2+\text{bz}_1)(\text{a}\bar{\text{z}_2}+\text{b}\bar{\text{z}_1})$ $=(\text{a}^2\text{z}_1\bar{\text{z}_1}-\text{ab}\text{z}_1\bar{\text{z}_2}-\text{ab}\text{z}_2\bar{\text{z}_1}+\text{b}^2\text{z}_2\bar{\text{z}_2})+(\text{a}^2\text{z}_2\bar{\text{z}_2}+\text{ab}\text{z}_1\bar{\text{z}_2}+\text{ab}\text{z}_2\bar{\text{z}_1}+\text{b}^2\text{z}_1\bar{\text{z}_1})$ $=[(\text{a}^2+\text{b}^2)\text{z}_1\bar{\text{z}_1}+(\text{a}^2+\text{b}^2)\text{z}_2\bar{\text{z}_2}]$ $=[(\text{a}^2+\text{b}^2)(\text{z}_1\bar{\text{z}_1}+\text{z}_2\bar{\text{z}_2})]$ $=[(\text{a}^2+\text{b}^2)(|\text{z}_1|^2+|\text{z}_2|^2)]$ Hence, $|\text{az}_1-\text{bz}_2|^2+|\text{az}_2+\text{bz}_1|^2=(\text{a}^2+\text{b}^2)\big(|\text{z}_1|^2+|\text{z}_2|^2\big).$
View full question & answer→Question 211 Mark
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ find the value of $\text{x}^2+\text{y}^2.$
Answer$\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy} \ ...(1)$ $\Rightarrow\Big[\overline{\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}}\Big]=\overline{\text{x}+\text{iy}}$ $\Rightarrow\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=\text{x}-\text{iy} \ ...(2)$ On multiplying (1) and (2), we get $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}\times\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=(\text{x}+\text{iy})(\text{x}-\text{iy})$ $\Rightarrow\frac{(\text{a}^2+1)^4}{(2\text{a})^2-\text{i}^2}=\text{x}^2-\text{i}^2\text{y}^2$ $\Rightarrow\frac{(\text{a}^2+1)^4}{(2\text{a})^2+1}=\text{x}^2+\text{y}^2$ Hence, $\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}.$
View full question & answer→Question 221 Mark
If $\pi<\theta<2\pi$ and $\text{z}=1+\cos\theta+\text{i}\sin\theta,$ then write the value of $|\text{z}|.$
Answer$\pi<\theta<2\pi$ $\frac{\pi}{2}<\frac{\theta}{2}<\pi$ (Dividing by 2) $\text{z}=1+\cos\theta+\text{i}\sin\theta$ $\Rightarrow|\text{z}|=\sqrt{(1+\cos\theta)^2+\sin ^2\theta}$ $\Rightarrow|\text{z}|=\sqrt{1+\cos^2\theta+2\cos\theta+\sin ^2\theta}$ $\Rightarrow|\text{z}|=\sqrt{1+1+2\cos\theta}$ $\Rightarrow|\text{z}|=\sqrt{2(1+2\cos\theta)}$ $\Rightarrow|\text{z}|=\sqrt{2\times2\cos^2\frac{\theta}{2}}$ $\Rightarrow|\text{z}|=2\sqrt{\cos^2\frac{\theta}{2}}$ $\Rightarrow|\text{z}|=-2\cos\frac{\theta}{2} \ \big[\text{Since}\frac{\pi}{2}<\frac{\theta}{2}<\pi,\cos\frac{\theta}{2} \ \text{is negative}\Big]$
View full question & answer→Question 231 Mark
If $|\text{z}|=2$ and $\text{arg(z)}=\frac{\pi}{4},$ find z.
AnswerWe know that, $\text{z}=|\text{z}|\big\{\cos[\text{arg(z)}]+\text{i}\sin[\text{arg(z)}]\big\}$ $=2\Big(\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{4}\Big)$ $=2\Big(\frac{1}{\sqrt{2}}+\text{i}\frac{1}{\sqrt{2}}\Big)$ $=\sqrt{2}(1+\text{i})$ Hence, $\text{z}=\sqrt{2}(1+\text{i}).$
View full question & answer→Question 241 Mark
Write the conjugate of $\frac{2-\text{i}}{(1-2\text{i})^2}.$
Answer$\frac{2-\text{i}}{(1-2\text{i})^2}=\frac{2-\text{i}}{1+4\text{i}^2-4\text{i}}$ $=\frac{2-\text{i}}{1-4-4\text{i}}$ $=\frac{2-\text{i}}{-3-4\text{i}}$ $=\frac{-2+\text{i}}{3+4\text{i}}$ $=\frac{\text{i}-2}{3+4\text{i}}\times\frac{3-4\text{i}}{3-4\text{i}}$ $=\frac{3\text{i}-4\text{i}^2-6+8\text{i}}{3^2-4^2\text{i}^2}$ $=\frac{11\text{i}+4-6}{9+16}$ $=\frac{-2}{25}+\frac{11}{25}\text{i}$ $\therefore$ Conjugate of $\frac{2-\text{i}}{(1-2\text{i})^2}=\Big(\overline{-\frac{2}{25}+\frac{11}{25}\text{i}}\Big)=-\frac{2}{25}-\frac{11}{25}\text{i}$ Hence, Conjugate of $\frac{2-\text{i}}{(1-2\text{i})^2}$ is $-\frac{2}{25}-\frac{11}{25}\text{i}.$
View full question & answer→Question 251 Mark
Find z, if $|\text{z}|=4$ and $\text{arg(z)}=\frac{5\pi}{6}.$
AnswerWe know that, $\text{z}=|\text{z}|\big\{\cos[\text{arg(z)]}+\text{i}\sin[\text{arg(z)}]\big\}$ $\Rightarrow\text{z}=4\Big(\cos\frac{5\pi}{6}+\text{i}\sin\frac{5\pi}{6}\Big)$ $=4\Big(-\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$ $=4\Big(-\frac{\sqrt{3}}{2}+\frac{1}{2}\text{i}\Big)$ $=-2\sqrt{3}+2\text{i}$ Thus, $\text{z}=-2\sqrt{3}+2\text{i}.$
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