Question

The number of complex number $z$ such that $\arg \left(z+3i\right)=\frac{\pi }{4}$ and $\arg \left(z-2\right)=\frac{\pi }{4}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$alg(z+3i)=\frac{\pi }{4}arg(z-2)=\frac{\pi }{4}$

Let $z=x+iy$

$\frac{y+3}{x}=\frac{y}{x-2}=1\left[\tan \frac{\pi }{4}\right]$

$y=x-2$

$\frac{x-2+3}{x}=1(puttingin\frac{y+3}{x}=1)$

$x+1=x$

$1=0-$contradiction

so there are no such complex numbers

$A$ is correct.