complex numbers

**jackm7** · March 17th 2009, 04:07 PM

hey, i have another question, this time to do with complex numbers.

it asks to find the real part of z=e^(1-i pi/3)

the way i started doing it was breaking it into e^1 - e^i pi/3
which gave me e^1 - cos pi/3 + sin pi/3
which reduces down to (1+sqrt(3))/2 + i/2) + e^1 after using special triangles.

does that make sense?

**TheEmptySet** · March 17th 2009, 04:14 PM

Originally Posted by jackm7

hey, i have another question, this time to do with complex numbers.

it asks to find the real part of z=e^(1-i pi/3)

the way i started doing it was breaking it into e^1 - e^i pi/3
which gave me e^1 - cos pi/3 + sin pi/3
which reduces down to (1+sqrt(3))/2 + i/2) + e^1 after using special triangles.

does that make sense?

Your Idea is good, but I think that

$e^{1-\frac{i\pi}{3}}=e^{1}\cdot e^{-\frac{\pi}{3}i}=e(\cos(\pi/3)-i\sin(\pi/3))$

Then just simplify and take the real part

**Scott H** · March 17th 2009, 04:18 PM

The formulas we can use here are:

$ \begin{aligned} a^{x+y} &= a^xa^y \\ e^{ix} &= \cos x+i \sin x.\\ \end{aligned} $

Therefore,

$ \begin{aligned} e^{1-i\frac{\pi}{3}}&=e^1e^{-i\frac{\pi}{3}}\\ &=e\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right)\\ &=e\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\\ &=\frac{1}{2}e-ie\frac{\sqrt{3}}{2}.\\ \end{aligned} $

Edit: Corrected the sign of the second exponent.

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