MCQ Complex Numbers and Quadratic Equations -

The value of $(\frac{1+i}{\sqrt{2}})^8+(\frac{1-i}{\sqrt{2}})^8$ is equal to
(A) 4
(B) 6
(C) 8
(D) 2

Answer:(D)
Explanation:
We have
$(\frac{1+i}{\sqrt{2}})^8+(\frac{1-i}{\sqrt{2}})^8$
$=[\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}]^8+[\cos \frac{\pi}{4}-i\sin \frac{\pi}{4}]^8$
$=[\cos 2\pi+i\sin 2\pi]+[\cos 2\pi-i\sin 2\pi]$ [by De-Moivre’s theorem]
$=2\cos 2\pi=2(1)=2$

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The solution of the equation $|z|-z=1+2i$ is
(A) $\frac{3}{2}-2i$
(B) $\frac{3}{2}+2i$
(C) $2-\frac{3}{2}i$
(D) $2+\frac{3}{2}i$

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Answer:(A)
Explanation:
Let $z=x+iy$
$|z|-z=1+2i$
$\Rightarrow \sqrt{x^2+y^2}-(x+iy)=1+2i$
$\Rightarrow \sqrt{x^2+y^2}-x-iy=1+2i$
Comparing real and imaginary parts
we have $\sqrt{x^2+y^2}-x=1$ and $y=-2$
Now substitute the value of $y$ in $\sqrt{x^2+y^2}-x=1$ we get
$\sqrt{x^2+4}-x=1$
$\Rightarrow \sqrt{x^2+4}=x+1$
$\Rightarrow x^2+4=(x+1)^2$
$\Rightarrow x^2+4=x^2+2x+1$
$\Rightarrow 4=2x+1$
$\Rightarrow x=\frac{3}{2}$

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If $\frac{z-1}{z+1}$ is purely imaginary, then
(A) $|z|>1$
(B) $|z|<1$
(C) $|z|=1$
(D) None of these

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Answer:(C)
Explanation:
Let $\frac{z-1}{z+1}=iy$ , where $y is real$
$\Rightarrow \frac{z+1}{z-1}=\frac{1}{iy}$
As we know by componendo and dividendo that $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}$
So we have
$\frac{z+1}{z-1}=\frac{1}{iy}\Leftrightarrow \frac{(z+1)+(z-1)}{(z+1)-(z-1)}=\frac{1+iy}{1-iy}$
$\Rightarrow \frac{2z}{2}=\frac{1+iy}{1-iy}$
$\Rightarrow z=\frac{1+iy}{1-iy}$
$\Rightarrow |z|=\frac{\sqrt{1+y^2}}{\sqrt{1+y^2}}=1$

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