Prove Complex Conjugate: z=cisθ

_wolfgang_
Messages
23
Reaction score
0

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...
 
Physics news on Phys.org
\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!
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top