Conjugate e^(iz): Solving for z = nπ

  • Thread starter guildmage
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In summary, the problem with Euler's formula is that it is valid only for real numbers. If z is a complex number, you can't use the cosine and sinine relations.
  • #1
guildmage
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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.

Homework Equations





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?
 
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  • #2
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
 
  • #3
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]
 
  • #4
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
 
  • #5
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])
 
  • #6
guildmage said:
[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]?
 
  • #7
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]
 
  • #8
Why even use sine?
exp(i z*)=exp(-i z*)
exp(2i z*)=1
z=n pi
 

What is the formula for solving for z in e^(iz) = nπ?

The formula for solving for z in e^(iz) = nπ is z = nπ/i, where i is the imaginary unit.

What does the conjugate of e^(iz) mean?

The conjugate of e^(iz) is the complex conjugate of the expression, which means changing the sign of the imaginary component. In this case, it would be e^(-iz).

Why is it important to solve for z in e^(iz) = nπ?

Solving for z allows us to find the values of z that satisfy the equation and can help us determine the values of n that make the equation true. It is important in many areas of mathematics and physics.

Can the equation e^(iz) = nπ have multiple solutions for z?

Yes, the equation can have multiple solutions for z. Since e^(iz) is a periodic function with a period of 2π, there can be infinitely many values of z that satisfy the equation for a given value of n.

How can the equation e^(iz) = nπ be used in real-world applications?

The equation e^(iz) = nπ can be used in various applications, such as in the study of oscillations and waves, signal processing, and electrical engineering. It also has applications in quantum mechanics and in solving differential equations.

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