# Complex numbers

by pamparana
Tags: complex, numbers
 P: 127 I just wanted to check something. If I have a complex number of the form $a = C * \exp(i \phi)$ where C is some non-complex scalar constant. Then the phase of this complex number is simply $\phi$. Is that correct?
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PF Gold
P: 3,288
 Quote by pamparana I just wanted to check something. If I have a complex number of the form $a = C * \exp(i \phi)$ where C is some non-complex scalar constant. Then the phase of this complex number is simply $\phi$. Is that correct?
If ##C > 0## then this is almost correct. However, the phase is not well-defined under this definition, because ##C\exp(i\phi) = C\exp(i(\phi+2\pi n))## for any integer ##n##. You can get around this by defining the phase to be the coset ##\phi + 2\pi \mathbb{Z}## or by constraining it to be in the interval ##[0,2\pi)## or ##[-\pi, \pi)## or some other half-open interval of length ##2\pi##.

If ##C < 0##, then you need to absorb the sign of ##C## into the phase:
$$a = -|C|\exp(i \phi) = |C|\exp(i(\phi + \pi))$$

If ##C = 0## then the phase is undefined.
 P: 127 Thank you for this detailed answer!

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