# Prove Complex Conjugate: z=cisθ

• _wolfgang_
In summary, the problem is to prove that for a complex number z=cis(theta), the conjugate is equal to 1/z. The attempt at a solution involves using a random complex number and trying to prove that 1/z is equal to 1 over the complex conjugate of z. However, this is incorrect. The correct way to solve the problem is by rationalizing the denominator and multiplying both numerator and denominator by the complex conjugate of z.

## Homework Statement

i am supposed to prove that for the complex number z=cis$$\theta$$
the conjugate is $$\frac{1}{\overline{z}}$$

## Homework Equations

if
z=a+bi
$$\overline{z}$$=a-bi

## The Attempt at a Solution

all that i can think of is that $$\frac{1}{cos\theta i sin \theta}$$
=(cos $$\theta$$ i sin $$\theta$$)-1

i have also just tried it with a random complex number such as w=2+3i
still how does $$\overline{w}$$=1/2+3i ?

Im very lost...

$$\frac{1}{\overline z}$$
is NOT equal to
$$\frac{1}{cos(\theta)isin(\theta)}$$

Also your problem, as stated, is wrong- the complex conjugate of $z= cis(\theta)$ is not 1 over the complex conjugate of z.
$$\frac{1}{\overline{z}= z$$
or
$$\frac{1}{z}= \overline{z}$$.

The complex conjugate of $cis(\theta)= cos(\theta)+ i sin(\theta)$ is $cos(\theta)- i sin(\theta)$.

The reciprocal of that is, of course,
$$\frac{1}{cos(\theta)- i sin(\theta)}$$

Now, "rationalize the denominator"- multiply both numerator and denominator by $cos(\theta)+ i sin(\theta)$

Conversely,
$$\frac{1}{z}= \frac{1}{cos(\theta)+ i sin(\theta)}$$

Multiply both numerator and denominator by $cos(\theta)- i sin(\theta)$.

Last edited by a moderator:
Ah that makes it a lot easyer, i totally forgot about when the denominator is imaginary that you multiply both numerator and denomiator by the conjugate.
Thanks alot!