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

Question

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

Accepted Answer

Let

$z=x+iy$

where $x,y\in R$.

We are given:

$|z+2|=|z-2|$

and

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

We will use these one by one.

1. Using $|z+2|=|z-2|$
Substitute $z=x+iy$:

$|x+iy+2|=|x+iy-2|$

So,

$\sqrt{(x+2{)}^{2}+{y}^{2}}=\sqrt{(x-2{)}^{2}+{y}^{2}}$

Squaring both sides:

$(x+2{)}^{2}+{y}^{2}=(x-2{)}^{2}+{y}^{2}$

${x}^{2}+4x+4={x}^{2}-4x+4$

$8x=0$

$x=0$

Hence,

$z=iy$

So the point lies on the imaginary axis.

2. Using the argument condition
Now,

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

We are given that its argument is $\frac{\pi }{4}$.

A complex number has argument $\frac{\pi }{4}$ when its real and imaginary parts are equal and positive in sign together.

Let us simplify:

$\frac{3+iy}{i(y-1)}$

Since $\frac{1}{i}=-i$, we get

$\frac{3+iy}{i(y-1)}=\frac{-i(3+iy)}{y-1}$

Now expand:

$-i(3+iy)=-3i-{i}^{2}y=y-3i$

So,

$\frac{z+3}{z-i}=\frac{y-3i}{y-1}=\frac{y}{y-1}-i\frac{3}{y-1}$

Thus real part is

$\frac{y}{y-1}$

and imaginary part is

$-\frac{3}{y-1}$

Since argument is $\frac{\pi }{4}$, we need

$\frac{}{}=	an \frac{\pi }{4}=1$

So,

$\frac{-\frac{3}{y-1}}{\frac{y}{y-1}}=1$

$-\frac{3}{y}=1$

$y=-3$

Therefore,

$z=-3i$

3. Find $|z{|}^{2}$
$|z{|}^{2}={0}^{2}+(-3{)}^{2}=9$

So,

$9$

Hence, the correct option is Option A.

Solutions · 1