Prove Complex Conjugate: z=cisθ

[_wolfgang_]

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...

 

[HallsofIvy]

$$\frac{1}{\overline z}$$
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 $cos(\theta)- i sin(\theta)$.

 

[_wolfgang_]

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!