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The phase of a complex number 
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#1
Feb1912, 08:59 PM

P: 6

in cartesian form, a+ ib you can find the phase by doing arctan(b/a).. my question concerns the phase of a purely imaginary number. during a lecture my professor said that the phase of i*2pi= pi/2, he rationalized this by saying that the number lies on the yaxis so the angle between the real axis and the imaginary axis is pi/2. but if you do arctan(2pi/0) you will get an error.. how is he right?



#2
Feb1912, 09:13 PM

Sci Advisor
P: 820

Complex numbers don't have a "phase" they have an "argument" which is defined case by case: if z = x + iy we define [itex]\phi = Arg(z)[/itex] as
arctan(y/x) when x > 0 arctan(y/x)+π when x < 0 and y ≥ 0 arctan(y/x)π when x < 0 and y < 0 π/2 when x = 0 and y > 0 π/2 when x = 0 and y < 0 indeterminate when x = 0 and y = 0. 


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