# Math Help - complex number

1. ## complex number

If z is a complex number such that modulus of z is 1 , find the real part of $\frac{1}{1-z}$

2. Hello,
Originally Posted by thereddevils
If z is a complex number such that modulus of z is 1 , find the real part of $\frac{1}{1-z}$
If the modulus of z is 1, this means that $z\bar z=1$, where $\bar z$ is the complex conjugate of z ( $\overline{x+iy}=x-iy$)

Property :
$\forall a\in\mathbb{R} ~,~ \overline{z+a}=\bar z+a$
(that's because the conjugate of a sum is the sum of the conjugates, and because the conjugate of a real number is itself)

Thus the conjugate of $1-z$ is $1-\bar{z}$

Multiply by the conjugate of the denominator :

$\frac{1-\bar{z}}{(1-z)(1-\bar z)}=\frac{1-\bar z}{1+z\bar z-(z+\bar z)}=\frac{1-\bar z}{2-2 \Re (z)}$

( $\Re$ denotes the real part)

so finally, the real part of $\frac{1}{1-z}$ is $\frac{1-\Re(z)}{2(1-\Re(z))}=\frac 12$

Looks good to you ?

3. Originally Posted by thereddevils
If z is a complex number such that modulus of z is 1 , find the real part of $\frac{1}{1-z}$
If $z = x + iy$ and $|z| = 1$, then

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

$x^2 + y^2 = 1$

Let's keep that in mind...

Now let's work on the $\frac{1}{1 - z}$.

$\frac{1}{1 - z} = \frac{1}{1 - x - iy}$

$= \frac{1}{1-x-iy}\times\frac{1-x + iy}{1 - x + iy}$

$= \frac{1 - x + iy}{(1-x)^2 + y^2}$

$= \frac{1 - x}{(1 - x)^2 + y^2} + i\frac{y}{(1 - x)^2 + y^2}$.

So the real part of $\frac{1}{1 - z}$ is $\frac{1 - x}{(1 - x)^2 + y^2}$, provided that $x^2 + y^2 = 1$

$= \frac{1 - x}{1 - 2x + x^2 + y^2}$

$= \frac{1 - x}{1 - 2x + 1}$ since $x^2 + y^2 = 1$

$= \frac{1 - x}{2 - 2x}$

$= \frac{1 - x}{2(1 - x)}$

$= \frac{1}{2}$.

So the Real part of $\frac{1}{1 - z}$ is $\frac{1}{2}$.