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Complex conjugate

  • Thread starter _wolfgang_
  • Start date
  • #1
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Homework Statement


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


Homework Equations


if
z=a+bi
[tex]\overline{z}[/tex]=a-bi

The Attempt at a Solution


all that i can think of is that [tex]\frac{1}{cos\theta i sin \theta}[/tex]
=(cos [tex]\theta[/tex] i sin [tex]\theta[/tex])-1

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

Im very lost.....
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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[tex]\frac{1}{\overline z}[/tex]
is NOT equal to
[tex]\frac{1}{cos(\theta)isin(\theta)}[/tex]

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

The complex conjugate of [itex]cis(\theta)= cos(\theta)+ i sin(\theta)[/itex] is [itex]cos(\theta)- i sin(\theta)[/itex].

The reciprocal of that is, of course,
[tex]\frac{1}{cos(\theta)- i sin(\theta)}[/tex]

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

Conversely,
[tex]\frac{1}{z}= \frac{1}{cos(\theta)+ i sin(\theta)}[/tex]

Multiply both numerator and denominator by [itex]cos(\theta)- i sin(\theta)[/itex].
 
Last edited by a moderator:
  • #3
23
0
Ah that makes it alot easyer, i totally forgot about when the denominator is imaginary that you multiply both numerator and denomiator by the conjugate.
Thanks alot!!
 

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