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 a lot!