If z is a complex number such that modulus of z is 1 , find the real part of $\displaystyle \frac{1}{1-z}$
Hello,
If the modulus of z is 1, this means that $\displaystyle z\bar z=1$, where $\displaystyle \bar z$ is the complex conjugate of z ($\displaystyle \overline{x+iy}=x-iy$)
Property :
$\displaystyle \forall a\in\mathbb{R} ~,~ \overline{z+a}=\bar z+a$
(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 $\displaystyle 1-z$ is $\displaystyle 1-\bar{z}$
Multiply by the conjugate of the denominator :
$\displaystyle \frac{1-\bar{z}}{(1-z)(1-\bar z)}=\frac{1-\bar z}{1+z\bar z-(z+\bar z)}=\frac{1-\bar z}{2-2 \Re (z)}$
($\displaystyle \Re$ denotes the real part)
so finally, the real part of $\displaystyle \frac{1}{1-z}$ is $\displaystyle \frac{1-\Re(z)}{2(1-\Re(z))}=\frac 12$
Looks good to you ?
If $\displaystyle z = x + iy$ and $\displaystyle |z| = 1$, then
$\displaystyle \sqrt{x^2 + y^2} = 1$
$\displaystyle x^2 + y^2 = 1$
Let's keep that in mind...
Now let's work on the $\displaystyle \frac{1}{1 - z}$.
$\displaystyle \frac{1}{1 - z} = \frac{1}{1 - x - iy}$
$\displaystyle = \frac{1}{1-x-iy}\times\frac{1-x + iy}{1 - x + iy}$
$\displaystyle = \frac{1 - x + iy}{(1-x)^2 + y^2}$
$\displaystyle = \frac{1 - x}{(1 - x)^2 + y^2} + i\frac{y}{(1 - x)^2 + y^2}$.
So the real part of $\displaystyle \frac{1}{1 - z}$ is $\displaystyle \frac{1 - x}{(1 - x)^2 + y^2}$, provided that $\displaystyle x^2 + y^2 = 1$
$\displaystyle = \frac{1 - x}{1 - 2x + x^2 + y^2}$
$\displaystyle = \frac{1 - x}{1 - 2x + 1}$ since $\displaystyle x^2 + y^2 = 1$
$\displaystyle = \frac{1 - x}{2 - 2x}$
$\displaystyle = \frac{1 - x}{2(1 - x)}$
$\displaystyle = \frac{1}{2}$.
So the Real part of $\displaystyle \frac{1}{1 - z}$ is $\displaystyle \frac{1}{2}$.