1. ## complex numbers

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?

2. 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

3. 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.