MATHS M.C.Q (1 Marks)

4. $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$

Solution:

 $(\text{i}^{25})^3=(\text{i})^{75}$

$=(\text{i})^{4\times18+3}$

$=(\text{i})^3$

$=-\text{i} \ (\because\text{i}^4=1)$

Let $\text{z}=0-\text{i}$

Since, the point (0,−1) lies on the negative direction of imaginary axis.Therefore, $\text{arg(z)}=\frac{-\pi}{2}$

Modulus, $\text{r}=|\text{z}|=|1|=1$

$\therefore$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$

$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$

$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$

4. $240^\circ$

Solution:

$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$

Rationalising the denominator,

$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$

$=\frac{1+3\text{i}^2-2\sqrt{3}\text{i}}{1-3\text{i}^2}$

$=\frac{-2-2\sqrt{3}\text{i}}{4} \ (\because\text{i}^2=-1)$

$=\frac{-1}{2}-\text{i}\frac{\sqrt{3}}{2}$

Then, $\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\sqrt{3}$

$\Rightarrow\alpha=60^\circ$

Since the points $(\frac{-1}{2},-\frac{-\sqrt{3}}{2})$ lie in the third quadrant, the argument is given by:

$\theta=180^\circ+60^\circ$

$=240^\circ$

3. $\text{i}\cot\frac{\theta}{2}$

​​​​​​​Solution:

$\text{a}=\cos\theta+\text{i}\sin\theta$ (given)

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-\text{i}\sin^2\theta+2\text{i}\sin\theta-\cos^2\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-(\sin^2\theta+\cos^2\theta)+2\text{i}\sin\theta}{1+(\sin^2\theta+\cos^2\theta)-2\cos\theta}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\theta}{2(1-\cos\theta)}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\frac{\theta}{2}-\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{\text{i}\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$

$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\text{i}\cot\frac{\theta}{2}$

2. The x-axis

Solution:

$\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$

$\Rightarrow \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|^2=1^2$

$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\overline{\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)}=1$

$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\Big(\frac{-\text{i}+\overline{\text{z}}}{-\text{i}-\overline{\text{z}}}\Big)=1$

$\Rightarrow\Big(\frac{-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}{-\text{i}^2+\text{z}\text{i}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}\Big)=1$

$\Rightarrow-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}=-\text{i}^2+\text{zi}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}$

$\Rightarrow-\text{zi}+\overline{\text{z}}\text{i}=\text{zi}-\overline{\text{z}}\text{i}$

$\Rightarrow\overline{\text{z}}\text{i}+\overline{\text{z}}\text{i}=\text{zi}-\text{zi}$

$\Rightarrow2\overline{\text{z}}\text{i}=2\text{zi}$

$\Rightarrow\overline{\text{z}}=\text{z}$

$ \Rightarrow\text{z}$ is purely real.

2. $|\text{z}|^2=|\text{z}|^2$

Solution:

It is obvious that, for any complex number z,

$|\text{z}|^2=|\text{z}|^2$

3. $2\Big|\sin\frac{\theta}{2}\Big|$

Solution:

$\therefore\text{z}=1-\cos\theta+\text{i}\sin\theta$

$\Rightarrow|\text{z}|=\sqrt{(1-\cos\theta)^2+\sin^2\theta}$

$\Rightarrow|\text{z}|=\sqrt{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$

$\Rightarrow|\text{z}|=\sqrt{1+1-2\cos\theta}$

$\Rightarrow|\text{z}|=\sqrt{2(1-2\cos\theta)}$

$\Rightarrow|\text{z}|=\sqrt{4\sin^2\frac{\theta}{2}}$

$\Rightarrow|\text{z}|=2\Big|\sin\frac{\theta}{2}\Big|$

3. 100

Solution:

$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$

Taking modulus on both the sides:

$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$

$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$

$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$

$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$

$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$

Squaring both the sides,

$\Rightarrow\text{x}^2+\text{y}^2=100$

1. $\pi$

Solution:

Given:

$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number

On rationalising, we get,

$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}\times\frac{{1+2\text{i}\sin\theta}}{{1+2\text{i}\sin\theta}}$

$=\frac{(3+2\text{i}\sin\theta)({1+2\text{i}\sin\theta})}{(1)^2-(2\text{i}\sin\theta)^2}$

$=\frac{3+2\text{i}\sin\theta+{6\text{i}\sin\theta}+4\text{i}^2\sin^2\theta}{1+4\sin^2\theta}$

$=\frac{3-4\text{i}\sin^2\theta+{8\text{i}\sin\theta}}{1+4\sin^2\theta} \ [\because\text{i}^2=-1]$

$=\frac{3-4\text{i}\sin^2\theta}{1+4\sin^2\theta}+\text{i}\frac{8\sin\theta}{1+4\sin^2\theta}$

For the above term to be real, the imaginary part has to be zero.

$\therefore\frac{8\sin\theta}{1+4\sin^2\theta}=0$

$\Rightarrow8\sin\theta=0$

For this to be zero,

$\sin\theta=0$

$\Rightarrow\theta=0,\pi,2\pi,3\pi...$

But $0<\theta<2\pi$

Hence, $\theta=\pi$

3. $\frac{2\pi}{3}$

​​​​​​​Solution:

$\text{z}=\frac{-2}{1+\sqrt{3}}$

Rationalising z, we get,

$\text{z}=\frac{-2}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$

$\Rightarrow\text{z}=\frac{-2+\text{i}2\sqrt{3}}{1+3}$

$\Rightarrow\text{z}=\frac{-1+\text{i}\sqrt{3}}{2}$

$\Rightarrow\text{z}=\frac{-1}{2}+\frac{\text{i}\sqrt{3}}{2}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\sqrt{3}$

$\Rightarrow\alpha=\frac{\pi}{3}$

Since, z lies in the second quadrant.

Therefore, $\text{arg(z)}=\pi-\frac{\pi}{3}$

$=\frac{2\pi}{3}$

2. circle with centre (-1, 0) and radius 1

Solution:

$|\text{z}+1|=1$

$\Rightarrow|\text{z+1|}^2=1^2$

$\Rightarrow(\text{z}+1)\overline{(\text{z}+1)}=1$

$\Rightarrow(\text{z}+1)(\overline{z}+1)=1$

$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}+1=1$

$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$

Since, $\text{z}=\text{x}+\text{iy}$

$\therefore\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$

$\Rightarrow(\text{x}+\text{iy})(\text{x}-\text{iy})+\text{x}+\text{iy}+\text{x}- \text{iy}=0$

$\Rightarrow\text{x}^2+\text{y}^2+\text{2x}=0$

$\Rightarrow(\text{x}+1)^2+(\text{y}-0)^2=1^2$

which is the equation of a circle with the center $(-1,\ 0)$ and radius 1.

1. $\frac{|\text{z}|}{2}$

Solution:

$\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$

$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$

$=\frac{7-1+2\text{i}}{1-(1^2+2^2\text{i}^2+4\text{i})}$

 $=\frac{6-2\text{i}}{1-1+4-4\text{i}}$

 $=\frac{6-2\text{i}}{4-4\text{i}}$

 $=\frac{6-2\text{i}}{4-4\text{i}}\times\frac{4+4\text{i}}{4+4\text{i}}$

$=\frac{24+24\text{i}-8\text{i}-8\text{i}^2}{4^2-4^2\text{i}^2}$

 $=\frac{24+16\text{i}+8}{16+16}$

 $=\frac{32+16\text{i}}{32}$

$=1+\frac{1}{2}\text{i}$

since $\text{z}=1+2\text{i},$

$\because|\text{z}|=\sqrt{(1)^2+(2)^2}$

$=\sqrt{1+4}$

$=\sqrt{5}$

$\therefore|\text{f}\text{(z)|}=\sqrt{(1)^1+(\frac{1}{2})^2}$

$=\sqrt{1+\frac{1}{4}}$ 

$=\frac{\sqrt5}{2}$

$=\frac{|\text{z}|}{2}$

3. $\text{a}^2+\text{b}^2$

​​​​​​​Solution:

$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$

Taking modulus on both the sides, we get,

$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$

$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$

$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$

$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$

Squaring on both the sides, we get:

$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$

1. 1

Solution:

$\text{a}-\text{ib}=\frac{3-4\text{ix}}{3+4\text{ix}}$

$=\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}$

$=\frac{9+16\text{x}^2\text{i}^2-24\text{xi}}{9-16\text{x}^2\text{i}^2}$

$=\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}$

 $\Rightarrow\ |\text{a}-\text{ib}|^2=\Bigg|\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}\Bigg|^2$

$\Rightarrow\text{a}^2+\text{b}^2=\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}$

$=\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}$

$=\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}$

$=\frac{(9+16\text{x}^2)^2}{(9+16\text{x}^2)^2}$

$=1$

2. $\frac{1}{\sqrt{26}}$

Solution:

Let $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})}$

 $\Rightarrow\text{z}=\frac{1}{2+\text{i}-3\text{i}^2}$

$\Rightarrow\text{z}=\frac{1}{2+\text{i}+3}$

$\Rightarrow\text{z}=\frac{1}{5+\text{i}}\times\frac{5-\text{i}}{5-\text{i}}$

$\Rightarrow\text{z}=\frac{5-\text{i}}{25-\text{i}^2}$

$\Rightarrow\text{z}=\frac{5-\text{i}}{25+1}$

$\Rightarrow\text{z}=\frac{5-\text{i}}{26}$

$\Rightarrow\text{z}=\frac{5}{26}-\frac{\text{i}}{26}$

$\Rightarrow|\text{z}|=\sqrt{\frac{25}{676}+\frac{1}{676}}$

$\Rightarrow\text{z}=\frac{1}{\sqrt{26}}$

2. $\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$

Solution:

$\text{z}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\times\frac{\text{a}+\text{ib}}{\text{a}+\text{ib}}$

$\Rightarrow\text{z}=\frac{\text{a}^2+\text{i}^2\text{b}^2+2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$

$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2+2\text{abi}}{\text{a}^2+\text{b}^2}$

$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}+\text{i}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$

$\Rightarrow\text{Re(z)}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2},\text{Im(z)}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$

$\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$

Since, z lies in the first quadrant . Therefore,

$\text{arg(z)}=\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$

$\tan\theta=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$

4. $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$

Solution:

$\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6}$

$\Rightarrow\text{z}=\frac{1}{\sqrt{2}}+\frac{1}{2}\text{i}$

$\Rightarrow|\text{z}|=\sqrt{\Big(\frac{1}{\sqrt{2}}\Big)^2+\frac{1}{4}}$

$\Rightarrow|\text{z}|=\sqrt{\frac{1}{2}+\frac{1}{4}}$

$\Rightarrow|\text{z}|=\sqrt{\frac{3}{4}}$

$\Rightarrow|\text{z}|=\frac{\sqrt{3}}{2}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\frac{1}{\sqrt{2}}$

$\Rightarrow\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$

Since, the point z lies in the first quadrant.

Therefore, $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$

2. 0

Solution:

$(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$

$=(1+\text{i})(1-1)(1-\text{i})(1+1) \ \big(\because\text{i}^2=-1, \beta=-\text{i and} \ \text{i}^4=1\big)$

$=(1+\text{i})(0)(1-\text{i})(2)$

$=0$

1. ​​​$\frac{1}{2}(1+\text{i})$

​​​​​​​Solution:

$\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$

$=\frac{\text{i}-1-\text{i}+1+\text{i}}{1+\text{i}} \ [$ As, $\text{i}^5=\text{i},\text{i}^6=-1,\text{i}^7=-\text{i},\text{i}^8=1,\text{i}^9=\text{i}]$

$=\frac{\text{i}}{\text{i}+1}$

$=\frac{\text{i}}{\text{i}+1}\times\frac{\text{i}-1}{\text{i}-1}$

$=\frac{\text{i}(\text{i}-1)}{\text{i}^2-1}$

$=\frac{\text{i}^2-\text{i}}{-2}$

$=\frac{1}{2}(1+\text{i})$

4. $\text{x}-\text{iy}$

Solution:

$\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy}$

Squaring on both the sides, we get,

$\text{a}+\text{ib}=\text{x}^2+(\text{iy})^2+2\text{ixy}$

$\Rightarrow\text{a}+\text{ib}=(\text{x}^2-\text{y}^2)+2\text{ixy}$

$\therefore\text{a}=(\text{x}^2-\text{y}^2)$

and $\text{b}=2\text{xy}$

$\therefore\text{a}-\text{ib}=(\text{x}^2-\text{y}^2)-2\text{ixy}$

$\Rightarrow\text{a}-\text{ib}=\text{x}^2+\text{i}^2\text{y}^2-2\text{ixy} \ [\because\text{i}^2=-1]$

Taking square root on both the sides, we get:

$\sqrt{\text{a}-\text{ib}}=\text{x}-\text{iy}$

2. ​​​-2

​​​​​​​Solution:

$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$

$=\frac{\text{i}^{4\times148}+\text{i}^{4\times147+2}+\text{i}^{4\times147}+\text{i}^{4\times146+2}+\text{i}^{4\times146}}{\text{i}^{4\times145+2}+\text{i}^{4\times145}+\text{i}^{4\times144+2}+\text{i}^{4\times144}+\text{i}^{4\times143+2}}-1$ $[\because\text{i}^4=1$ and $\text{i}^2=-1]$

$=\frac{1+\text{i}^2+1+\text{i}^2+1}{\text{i}^2+1+\text{i}^2+1+\text{i}^2}-1$

$=\frac{1}{-1}-1$

$=-2$

2. 8

​​​​​​​Solution:

 Let $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)$

$\Rightarrow\text{z}=\frac{2\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$

$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1-\text{i}^2}$

$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1+1} \ [\because\text{i}^2=-1]$

$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{2}$

$\Rightarrow\text{z}=\text{i}-\text{i}^2$

$\Rightarrow\text{z}=\text{i}+1$

Now, $\text{z}^\text{n}=(1+\text{i})^\text{n}$

For $\text{n}=2,$

$\text{z}^2=(1+\text{i})^2$

$=1+\text{i}^2+2\text{i}$

$=1-1+2\text{i}$

$=2\text{i} \ ...(1)$

Since this is not a positive integer,

For $\text{n}=4,$

$\text{z}^4=(1+\text{i})^4$

$=\big[(1+\text{i})^2\big]^2$

$=(2\text{i})^2$ [Using (1)]

$=(4\text{i})^2$

$=-4 \ ...(2)$

This is a negative integer.

For $\text{n}=8,$

$\text{z}^8=(1+\text{i})^8$

$=\big[(1+\text{i})^4\big]^2$

$=(-4)^2$ [Using (2)]

$=16$

This is a positive integer.

Thus, $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is positive for $\text{n}=8.$

Therefore, 8 is the least positive integer such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer.

3. ​​​$-\frac{\pi}{2}$

​​​​​​​Solution:

Let $\text{z}=\frac{1}{\text{i}}$

$\text{z}=\frac{1}{\text{i}}\times\frac{\text{i}}{\text{i}}$

$\text{z}=\frac{\text{i}}{\text{i}^2}$

$\text{z}=-\text{i}$

Since, z (0, -1) lies on the negative imaginary axis . Therefore, $\text{arg(z)}=\frac{-\pi}{2}$

3. $\text{b}<\text{a}<0$

Solution:

Since, $\text{z}=\text{a}+\text{ib}$ lies in third quadrant.

$\Rightarrow\text{a}<0$ and $\text{b}<0 \ ...(1)$

Now,

$\frac{\bar{\text{z}}}{\text{z}}=\frac{\overline{\text{a}+\text{ib}}}{\text{a}+\text{ib}}$

$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$

$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}-\text{ib}}$

$=\frac{\text{a}^2+\text{i}^2\text{b}^2-2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$

$=\frac{\text{a}^2-\text{b}^2-2\text{abi}}{\text{a}^2+\text{b}^2}$

Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant.

$\Rightarrow\text{a}^2-\text{b}^2<0$

$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b})<0$

$\Rightarrow\text{a}-\text{b}>0$ and $\text{a}+\text{b}<0$

$\Rightarrow\text{a}>\text{b} \ ...(2)$

From (1) and (2),

$\text{b}<\text{a}<0$

1. 0

Solution:

$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$

$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$

$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$

$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$

$\Rightarrow\text{z}=1$

Since point (1,0) lies on the positive direction of real axis, we have:

arg(z) = 0

2. $-\sqrt{6}$

Solution:

$\sqrt{-2}\times\sqrt{-3}$

$=\sqrt{2}\times\sqrt{3}\times\sqrt{-1}\times\sqrt{-1}$

$=\sqrt{6}\times\text{i}\times\text{i}$

$=\sqrt{6}\times\text{i}^2$

$=-\sqrt{6} \ [\because\text{i}^2=-1]$

3. ​​​$\frac{\pi}{6}$

​​​​​​​Solution:

Let $\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$

$\Rightarrow\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}\times\frac{\sqrt{3}-\text{i}}{\sqrt{3}-\text{i}}$

$\Rightarrow\text{z}=\frac{\sqrt{3}+2\text{i}-\sqrt{3}\text{i}^2}{3-\text{i}^2}$

$\Rightarrow\text{z}=\frac{\sqrt{3}+\sqrt{3}+2\text{i}}{4}$

$\Rightarrow\text{z}=\frac{2\sqrt{3}+2\text{i}}{4}$

$\Rightarrow\text{z}=\frac{\sqrt{3}}{2}+\frac{1}{2}\text{i}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\frac{1}{\sqrt{3}}$

$\Rightarrow\alpha=\frac{\pi}{6}$

Since, z lies in the first quadrant. Therefore, $\text{arg(z)}=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=\frac{\pi}{6}$

2. $2\text{i}$

​​​​​​​Solution:

$\text{a}=1+\text{i}$

On squaring both the sides, we get,

$\text{a}^2=(1+\text{i})^2$

$\Rightarrow\text{a}^2=1+\text{i}^2+2\text{i}$

$\Rightarrow\text{a}^2=1-1+2\text{i} \ (\because\text{i}^2=-1)$

$\Rightarrow\text{a}^2=2\text{i}$

1. $\frac{\pi}{4}$

​​​​​​​Solution:

Let $\text{z}=(1+\text{i})$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=1$

$\Rightarrow\alpha=\frac{\pi}{4}$

Since, z lies in the first quadrant.

Therefore, $\text{arg(z)}=\frac{\pi}{4}$

4.  none of these.

Solution:

$(\text{x}+\text{iy)}^{\frac{1}{3}}=\text{a}+\text{ib}$

Cubing on both the sides, we get:

$\text{x}+\text{iy}=(\text{a}+\text{ib})^3$

$\Rightarrow\text{x}+\text{iy}=\text{a}^3+(\text{ib})^3+3\text{a}^2\text{bi}+3\text{a}(\text{bi})^2$

$\Rightarrow\text{x}+\text{iy}=\text{a}^3+\text{i}^3\text{b}^3+3\text{a}^2\text{bi}+3\text{i}^2\text{ab}^2$

$\Rightarrow\text{x}+\text{iy}=\text{a}^3-\text{i}\text{b}^3+3\text{a}^2\text{bi}-3\text{ab}^2 \ (\because\text{i}^2=-1,\text{i}^3=-\text{i})$

$\Rightarrow\text{x}+\text{iy}=\text{a}^3-3\text{a}\text{b}^2+\text{i}(-\text{b}^3+3\text{a}^2\text{b})$

$\therefore\text{x}=\text{a}^3-3\text{a}\text{b}^2$ and $\text{y}=3\text{a}^2\text{b}-\text{b}^3$

or, $\frac{\text{x}}{\text{a}}=\text{a}^2-3\text{b}^2$ and $\frac{\text{y}}{\text{b}}=3\text{a}^2-\text{b}^2$

$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\text{a}^2-3\text{b}^2+3\text{a}^2-\text{b}^2$

$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=4\text{a}^2-4\text{b}^2$

4. ​​​$\text{amp(z)}=\frac{3\pi}{4}$

​​​​​​​Solution:

$\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$

$\Rightarrow\text{z}=\frac{1+7\text{i}}{4-1-4\text{i}} \ [\because\text{i}^2=-1]$

$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}$

$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}$

$\Rightarrow\text{z}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{9-16\text{i}^2}$

$\Rightarrow\text{z}=\frac{3-28+25\text{i}}{9+16}$

$\Rightarrow\text{z}=\frac{-25+25\text{i}}{25}$

$\Rightarrow\text{z}=-1+\text{i}$

$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=1$

$\Rightarrow\alpha=\frac{\pi}{4}$

Since, z lies in the second quadrant.

Therefore, $\text{amp(z)}=\pi-\alpha$

$=\pi-\frac{\pi}{4}$

$=\frac{3\pi}{4}$

1. $\frac{1}{13}$

Solution:

Let $\text{z}=\frac{1}{(2+3\text{i})^2}$

$\Rightarrow\text{z}=\frac{1}{4+9\text{i}^2+12\text{i}}$

$\Rightarrow\text{z}=\frac{1}{4-9+12\text{i}}$

$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}$

$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}\times\frac{-5-12\text{i}}{-5-12\text{i}}$

$\Rightarrow\text{z}=\frac{-5-12\text{i}}{25+144}$

$\Rightarrow\text{z}=\frac{-5}{169}-\frac{12\text{i}}{169}$

$\Rightarrow|\text{z}|=\sqrt{\frac{25}{169^2}+\frac{144}{169^2}}$

$\Rightarrow|\text{z}|=\frac{1}{\sqrt{169}}$

$\Rightarrow|\text{z}|=\frac{1}{13}$

1. $|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$

Solution:

since we know that

$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$

$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)+\text{arg}(\text{z}_2)$ and

$|\text{z}_1+\text{z}_2|\le|\text{z}_1|+|\text{z}_2|$

1. $\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$

​​​​​​​Solution:

$\text{x}+\text{iy}=\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}$

Taking modulus on both the sides, we get:

$\sqrt{\text{x}^2+\text{y}^2}=\frac{(\text{a}^2+1)^2}{\sqrt{4\text{a}^2+1}}$

Squaring both sides, we get,

$\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{{4\text{a}^2+1}}$

1. ​​​$-\frac{\pi}{2}$

​​​​​​​Solution:

Let $\text{z}=\frac{1-\text{i}}{1+\text{i}}$

$\Rightarrow\text{z}=\frac{1-\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$

$\Rightarrow\text{z}=\frac{1+\text{i}^2-2\text{i}}{1-\text{i}^2}$

$\Rightarrow\text{z}=\frac{1-1-2\text{i}}{1+1}$

$\Rightarrow\text{z}=\frac{-2\text{i}}{2}$

$\Rightarrow\text{z}=\text{i}$

Since, z lies on negative direction of imaginary axis . Therefore, $\text{arg(z)}=\frac{-\pi}{2}$

3. -8

Solution:

Using $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$

$(1+\text{i})^4+(1-\text{i})^4$

$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$

$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$

$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$

$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$

$=8\text{i}^2$

$=-8$​​​​​​​

3. ​​​-i

​​​​​​​Solution:

Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$

$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$

$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$

$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$

$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$

$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$

$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$

$\Rightarrow\text{z}=-\text{i}$

1. $\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$

​​​​​​​Solution:

$\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big| \ \Big(\because\text{z}\bar{\text{z}}=|\text{z}|^2\Big)$

Let $\text{z}=\text{a}+\text{ib}$

$\Rightarrow|\text{z}|=\sqrt{\text{a}^2+\text{b}^2}$

Let $\bar{\text{z}}=\text{a}-\text{ib}$

$\Rightarrow|\bar{\text{z}}|=\sqrt{\text{a}^2+\text{b}^2}$

$\therefore\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big|$

$=\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$

1. 1

Solution:

Let $\text{z}=\frac{1+\text{i}}{1-\text{i}}$

Rationalising the denominator:

$\text{z}=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$

$\Rightarrow\text{z}=\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}$

$\Rightarrow\text{z}=\frac{1-1+2\text{i}}{1+1}$

$\Rightarrow\text{z}=\frac{2\text{i}}{2}$

$\Rightarrow\text{z}=\text{i}$

$\Rightarrow\text{z}^4=\text{i}^4$

Since $\text{i}^2=-1,$ we have:

$\Rightarrow\text{z}^4=\text{i}^2\times\text{i}^2$

$\Rightarrow\text{z}^4=1$

3. $\text{a}^2+\text{b}^2$

Solution:

$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$

Taking modulus on both the sides, we get:

$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=|\text{a}+\text{ib}|$

$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be written as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$

$\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$

$\Rightarrow\sqrt{2}\times\sqrt{5}\times\sqrt{10}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$

Squaring on both the sides, we get:

$2\times5\times10\times...\times(1+\text{n}^2)=\text{a}^2+\text{b}^2$

4. 0

Solution:

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}$

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i} \ \text{and} \ \text{i}^4=1]$

$=\text{i}-1-\text{i}+1$

$=0$

Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4. Hence, the sum of all terms, till 1000, will be zero.

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}=0$

1. 1

Solution:

$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$

Taking modulus on both the sides, we get:

$\Big|\frac{1-\text{ix}}{1+\text{ix}}\Big|=\big|\text{a}+\text{ib}\big|$

$\Rightarrow\frac{\sqrt{1^2+\text{x}^2}}{\sqrt{1^2+\text{x}^2}}=\sqrt{\text{a}^2+\text{b}^2}$

$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=1$

Squaring both the sides, we get:

$\text{a}^2+\text{b}^2=1$

2. $\frac{1}{2}$

Solution:

$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}$

$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$

$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$

$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+\cos^2\theta-2\cos\theta+\text{i}\sin^2\theta}$

$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+1-2\cos\theta}$

$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{2(1-\cos\theta)}$

$\Rightarrow\text{Re(z)}=\frac{(1-\cos\theta)}{2(1-\cos\theta)}=\frac{1}{2}$

3. $\frac{53}{85}$

Solution:

$\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}$

$\Rightarrow\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}\times\frac{7+6\text{i}}{7+6\text{i}}$

$\Rightarrow\text{x}+\text{iy}=\frac{21+53\text{i}+30\text{i}^2}{49-36\text{i}^2}$

$\Rightarrow\text{x}+\text{iy}=\frac{21-30+53\text{i}}{49+36}$

$\Rightarrow\text{x}+\text{iy}=\frac{-9}{85}+\text{i}\frac{53}{85}$

On comparing both the sides:

$\text{y}=\frac{53}{85}$