# Thread: solve for z

1. ## solve for z

$z$ is not equal to $i$ (didn't find the "not equal to" sign)

Show that

$\mid \frac{i - z}{1 - i\overline{z}}\mid = 1$

I tried several times by putting in z = a + bi.. without luck. can someone show me how to do this?

2. Originally Posted by metlx
$z$ is not equal to $i$ (didn't find the "not equal to" sign)

Show that

$\mid \frac{i - z}{1 - i\overline{z}}\mid = 1$

I tried several times by putting in z = a + bi.. without luck. can someone show me how to do this?

As you said is fine: $z=x+yi\Longrightarrow i-z=-x+(1-y)y\,,\,\,1-i\overline{z}=(1-y)-ix$

So you see the numerator's real parts equals the denominators imaginary part and the numerator's imaginary part equals the denominator's real part and thus the modulus of both denominator and denominator are equal...

Tonio

3. Originally Posted by metlx
$z$ is not equal to $i$ (didn't find the "not equal to" sign)

Show that

$\mid \frac{i - z}{1 - i\overline{z}}\mid = 1$

I tried several times by putting in z = a + bi.. without luck. can someone show me how to do this?

Problem: Compute $\left|\frac{i-z}{1-i\cdot\bar{z}}\right|$ given that $z\ne i$

Solution: Suppose that $z=a+bi$ then $\bar{z}=a-bi$ and $i\cdot\bar{z}=ai+b$. So that $\left|\frac{i-z}{-i\cdot\bar{z}}\right|=\left|\frac{-a-(b-1)i}{(1-b)-ai}\right|=\frac{\left|-a-(b-1)i\right|}{\left|(1-b)-ai\right|}=\frac{\sqrt{a^2+\left(b-1\right)^2}}{\sqrt{a^2+\left(b-1\right)^2}}=1$

EDIT: Oops, sorry was a little late.