Why do all numbers in the complex plane define a unit circle radius?

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rustynail
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Hi, I have recently started learning about complex numbers and the complex plane, and there is something I don't quite understand. Why is it that for every number [itex]z \in C[/itex]

[itex]e^{i\theta} = z[/itex]

we have

[itex]\sqrt { [Re(z)]² + [Im(z)]² } = 1[/itex] ?


Except for [itex]\theta = \frac {\pi}{2} or \frac {3\pi}{2}[/itex] of course. In other words, why do all numbers z define a unit circle's radius ?

Any help would be very much appreciated
 
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That is not true. Only complex numbers of the form z = e^(iθ) lie on the unit circle and satisfy that equation. Complex numbers of the form z = Re^(iθ) lie on circles of radius R in the complex plane, as you can verify.
 
I have stated that z must satisfy

rustynail said:
[itex]e^{i\theta} = z[/itex]
 
In that case, apply Euler's formula: e^(iθ) = cos(θ) + i*sin(θ), from which the result is immediate. The derivation of Euler's formula is implied by the Taylor series for e^x, cos(x) and sin(x), but its full justification requires a little more rigor.
 
Ok so adding the taylor series for cos(θ) and for i sin(θ) should give the taylor series for e^(iθ) ? I will definitely try that out. Thank you!
 
rustynail said:
Ok so adding the taylor series for cos(θ) and for i sin(θ) should give the taylor series for e^(iθ) ? I will definitely try that out. Thank you!

Yes, that's one of the proofs. Definitely worth going through the steps. Euler's formula falls right out of the power series for e^(iθ).
 
Also if you apply the fact that the complex conjugate of e^(iθ) = e^(-iθ) then multiply the two gives you e^(iθ - iθ) = e^(0) = 1.

You can prove this using the propery that cos(-x) = cos(x) and -sin(x) = sin(-x) using properties of odd and even functions.