Let z be a complex number such that |z+2|=|z-2| and (z+3z-i)=4.

Question

Let $z$ be a complex number such that $|z+2|=|z-2|$ and $\arg\left(\dfrac{z+3}{z-i}\right)=\dfrac{\pi}{4}$. Then $|z|^2$ is equal to:

Accepted Answer

Let $z=x+iy$. From $|z+2|=|z-2|$:
$$(x+2)^2+y^2=(x-2)^2+y^2 \implies 8x=0 \implies x=0.$$

So $z=iy$. Then:
$$\frac{z+3}{z-i}=\frac{3+iy}{i(y-1)}=\frac{-i(3+iy)}{y-1}=\frac{y-3i}{y-1}=\frac{y}{y-1}-\frac{3}{y-1}i.$$

Since the argument is $\dfrac\pi4$, real and imaginary parts must be equal (and positive):
$$\frac{-3/(y-1)}{y/(y-1)}=\tan\frac\pi4=1 \implies -\frac{3}{y}=1 \implies y=-3.$$

So $z=-3i$, and $|z|^2=0^2+(-3)^2=9$.

Solutions · 1