Question

(1) For any two complex numbers $z_1$, $z_2$ and any real numbers $a$, $b$, ${|a{z}_1-b{z}_2|}^2+{|b{z}_1+a{z}_2|}^2=$

(2) The value of $\sqrt{-25}\times \sqrt{-9}$ is

(3) The number $\frac{{(1-i)}^3}{1-i^3}$ is
equal to

(4) The sum of the series $i+i^2+i^3+......$ upto $1000$ terms is .

(5) Multiplicative inverse of $(1+i)$ is

(6) If $z_1$ and $z_2$ are complex numbers such that $z_1+z_2$ is real number then .

(7) $\text{arg}\left(z\right)+\text{arg}\left(\overline{z}\right)(\overline{z}\ne 0)$ is

(8) If $|z+4|\le 3$, then the greatest and least value of $|z+1|$ are and .

(9) If $\left|\frac{z-2}{z+2}\right|=\frac{\pi }{6}$, then the locus of $z$ is .

(10) If $\left|z\right|=4$ and $\text{arg}\left(z\right)=\frac{5\pi }{6}$, then $z=$

• A. 0
• B. -15
• C. $Im\left(z_1\right)+Im\left(z_2\right)=0$
• D. zero
• E. 0
• F. -2
• G. $z=-2\sqrt{3}+2i$
• H. 6
• I. $(a^2+b^2)({\left|z_1\right|}^2+{\left|z_2\right|}^2)$
• J. a circle
• K. $\frac{1}{2}-\frac{i}{2}$

Answer 1

Answer: (A), (B), (C), (D), (E), (F), (G), (H), (I), (J), (K)

(1)
Using the property,
${|z_1\pm z_2|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2\pm 2Re\left(z_1\overline{z_2}\right)$

$\therefore {|a{z}_1-b{z}_2|}^2+{|b{z}_1+a{z}_2|}^2$

$={\left|a{z}_1\right|}^2+{\left|b{z}_2\right|}^2-2Re(a{z}_1\times \overline{b{z}_2})+{\left|b{z}_1\right|}^2+{\left|a{z}_2\right|}^2+2Re(b{z}_1\times \overline{a{z}_2})$

$={\left|a{z}_1\right|}^2+{\left|b{z}_2\right|}^2-2Re\left(ab{z}_1\overline{z_2}\right)+{\left|b{z}_1\right|}^2+{\left|a{z}_2\right|}^2+2Re\left(ab{z}_1\overline{z_2}\right)$

$=(a^2+b^2){\left|z_1\right|}^2+(a^2+b^2){\left|z_2\right|}^2$

$=(a^2+b^2)({\left|z_1\right|}^2+{\left|z_2\right|}^2)$

Hence,
${|a{z}_1-b{z}_2|}^2+{|b{z}_1+a{z}_2|}^2=(a^2+b^2)({\left|z_1\right|}^2+{\left|z_2\right|}^2)$


(2)
$\because \sqrt{-25}=\sqrt{-1}\times \sqrt{25}$
$=\sqrt{25}i[\because i=\sqrt{-1}]$
$=5i$

$\sqrt{-9}=\sqrt{-1}\times \sqrt{9}$
$=\sqrt{9}i$
$=3i$

So, $\sqrt{-25}\times \sqrt{-9}=5i\times 3i$
$=15i^2$
$=-15[\because i^2=-1]$

Hence, $\sqrt{-25}\times \sqrt{-9}=-15$


(3)
$\frac{{(1-i)}^3}{1-i^3}=\frac{{(1-i)}^3}{1+i}[\because i^3=-i]$

$=\frac{1-i^3-3i(1-i)}{1+i}$
$[\because {(a-b)}^3=a^3-b^3-3ab(a-b)]$

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

$=\frac{1-3-2i}{1+i}[\because i^2=-1]$

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

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

$=-2$


(4)
We know,
$[i^n+i^{n+1}+i^{n+2}+i^{n+3}=0,\text{where}n\in N]$
To find: $i+i^2+i^3+......$ upto $1000$ terms
i.e. $i+i^2+i^3+...+i^{1000}$
$=(i+i^2+i^3+i^4)+(i^5+i^6+i^7+i^8)+...+(i^{997}+i^{998}+i^{999}+i^{1000})$
$=0+0+...+0$
$=0$
Hence, the value of the required sum is $0$.


(5)
Let $z=1+i$
Multiplicative inverse of $z=\frac{1}{z}$

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

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

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

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

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

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


(6)
Let $z_1=a+ib$, $z_2=x+iy$
Now,
$z_1+z_2=(a+x)+i(b+y)$

For $z_1+z_2$ is a real number, we get
$b+y=0$
$\Rightarrow Im\left(z_1\right)+Im\left(z_2\right)=0$


(7)
$\text{arg}\left(z\right)+\text{arg}\left(\overline{z}\right)$
$=\text{arg}\left(z\overline{z}\right)$
$[\because \text{arg}\left(z_1\right)+\text{arg}\left(z_2\right)=\text{arg}\left(z_1{z}_2\right)]$
$=\text{arg}\left({\left|z\right|}^2\right)[\because z\overline{z}={\left|z\right|}^2]$
$=0$ [$\because {\left|z\right|}^2$ is real and positive]


(8)
Given: $|z+4|\le 3$
$\because |z+1|=|z+4-3|$
$\Rightarrow |z+1|\le |z+4|+|-3|\{\because |a+b|\le \left|a\right|+\left|b\right|\}$
$\Rightarrow |z+1|\le 3+3\{\because |z+4|\le 3\}$
$\Rightarrow |z+1|\le 6$
So, the greatest value of $|z+1|$ is $6$
$\Rightarrow |z+1|\ge ||z+4|-\left|3\right||\ge 0$
So, the least value of $|z+1|$ is zero.
( Equality holds when $|z+4|=3$)


(9)
Given:$\left|\frac{z-2}{z+2}\right|=\frac{\pi }{6}z\ne -2$

Let $z=x+iy$

$\Rightarrow \left|\frac{x+iy-2}{x+iy+2}\right|=\frac{\pi }{6}$

$\Rightarrow \frac{|x-2+iy|}{|x+2+iy|}=\frac{\pi }{6}[\because \left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}]$

$\Rightarrow |x-2+iy|=\pi |x+2+iy|$
$\Rightarrow 6\sqrt{{(x-2)}^2+{\left(y\right)}^2}=\pi \sqrt{{(x+2)}^2+{\left(y\right)}^2}$

Squaring on both sides, we get

$\Rightarrow 36[{(x-2)}^2+y^2]=\pi [{(x+2)}^2+y^2]$

$\Rightarrow 36[x^2+4-4x+y^2]=\pi [x^2+4+4x+y^2]$

$\Rightarrow (36-{\pi }^2)x^2+(36-{\pi }^2)y^2-(144+4{\pi }^2+x)+4(36-{\pi }^2)=0$

$\Rightarrow x^2+y^2-\left(\frac{144+4{\pi }^2}{36-{\pi }^2}\right)x+4=0$
The above equation represents a circle.
Hence, the locus of $z$ is a circle.


(10)
Given: $\left|z\right|=4$ and $\text{arg}\left(z\right)=\frac{5\pi }{6}$
From polar form of a complex number, we get
$z=\left|z\right|(\cos \theta +i\sin \theta )$

$\Rightarrow z=4(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6})$

$\Rightarrow z=4(\cos (\pi -\frac{5\pi }{6})+i\sin (\pi -\frac{5\pi }{6}))$

$\Rightarrow z=4(-\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$

$\Rightarrow z=4(\frac{\sqrt{3}}{2}+i\frac{1}{2})$

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