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Conjugate of e^(iz)

  • Thread starter guildmage
  • Start date
1. Homework Statement

I want to show that

[tex] \exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)} [/tex]

if and only if

[tex] z = n\pi [/tex]

for any integer n.

2. Homework Equations



3. The Attempt at a Solution
Utilizing Euler's formula, I got

[tex] \cos \bar{z} = \cos z [/tex]

and

[tex] \sin \bar{z} = -\sin z [/tex]

Though not fully convinced, I concluded that

[tex] \bar{z} = z [/tex]

This then led me to

[tex] \sin z = 0 [/tex]

This obviously led me to the needed conclusion. But was I correct?
 

Office_Shredder

Staff Emeritus
Science Advisor
Gold Member
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98
Euler's formula is only valid for z a real number. Taking conjugates doesn't get you very far there. You should split z into x+iy and divvy up the exponential first
 
But Euler's formula has already been proven to apply even for complex numbers. My primary concern is whether or not

[tex]
\exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)}
[/tex]

would give me

[tex]
\bar{z} = z
[/tex]
 

Office_Shredder

Staff Emeritus
Science Advisor
Gold Member
3,734
98
I didn't mean valid there, I meant useful.

For example:

[tex] \cos \bar{z} = \cos z [/tex]

and

[tex] \sin \bar{z} = - \sin z [/tex]

is satisfied by any imaginary number (just by using the even/oddness of cosine/sine).

The problem though, is that if z is a complex number, you don't have that cos(z) and sin(z) are real numbers, so you can't draw the two relations you have
 

lurflurf

Homework Helper
2,417
122
exp(i z)*=exp((i z)*)=exp(i*z*)=exp(-i z*)
it is also easy to see
exp(i z)=exp(-Im[z]+i Re[z])=exp(-Im[z])exp(i Re[z])
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
[tex] \cos \bar{z} = \cos z [/tex]

and

[tex] \sin \bar{z} = -\sin z [/tex]

Though not fully convinced, I concluded that
[tex]
\bar{z} = z
[/tex]
That's a strange conclusion:wink:.....If I told you [itex]\cos(0)=\cos(100\pi)[/itex], would you then conclude [itex]0=100\pi[/itex]?
 
Alright. What if I say

[tex]
\cos \bar{z} = \cos z
[/tex]

would give me

[tex]
\bar{z} = z + 2n\pi
[/tex]

(Is this correct?)

I will then use it to say that

[tex]
\sin \bar{z} = -\sin z
[/tex]

gives me

[tex] \sin \left( z + 2n\pi \right) = -\sin z [/tex]

This implies that

[tex] \sin z = -\sin z [/tex]

Therefore,

[tex] z = n\pi [/tex]
 

lurflurf

Homework Helper
2,417
122
Why even use sine?
exp(i z*)=exp(-i z*)
exp(2i z*)=1
z=n pi
 

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