Question

The number of complex numbers $p$ such that $|p|=1$ and imaginary part of $p^4$ is $0$ , is

• A. $4$
• B. $2$
• C. $8$
• D. infinitely many

Answer 1

The correct option is C $8$

$|p|=1$----(1)

imaginary part of $p^4=0$

Let $P=x+iy$

$\Rightarrow p^2=x^2-y^2+2ixy$

$\Rightarrow p^4=(x^2-y^2{)}^2-4x^2{y}^2+4ixy(x^2-y^2)$

It is given that, imaginary part of $p^4$ is zero.

$\Rightarrow 4ixy(x^2-y^2)=0$

$\Rightarrow x^2-y^2=0$

$\Rightarrow x=\pm y$

Now from equation (1)

$|p|=1$

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

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

$\Rightarrow 2y^2=1$

$\Rightarrow y=\pm \frac{1}{\sqrt{2}}$

i.e., there are four complex numbers satisfies the given conditions.

(or)

Given:

$|p|=1$, imaginary part of $p^4=0$i.e circle with centre $(0,0)$ & radius $=1$

$A=(1,0)B=(0,1)C=(-1,0)D=(0-1)$
$E=1(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$

$\Rightarrow E^4=(\cos \frac{4\pi }{4}+i\sin \frac{4\pi }{4})$

$\Rightarrow E^4=\cos \pi +i\sin \pi F=(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})G=\cos (-\frac{3\pi }{4})+i\sin \left(\frac{-3\pi }{4}\right)H=\cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4})i.e4$