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

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

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}

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

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