Originally Posted by

**Showcase_22** The question is:

Let $\displaystyle z=\frac{\pi}{6}+iln2$. Write $\displaystyle e^{iz}$ in the form a+bi.

(careful! This is a trick question.)

What I did:

$\displaystyle e^{iz}=e^{\frac{i \pi}{6}-ln2}$

$\displaystyle e^{iz}=e^{\frac{i\pi}{6}}e^{-ln2}$

$\displaystyle e^{iz}=\frac {1}{2} e^{\frac{i\pi}{6}}$

$\displaystyle e^{iz}=(\frac{1}{2})(cos(\frac{\pi}{6}) + i sin(\frac{\pi}{6}))$

$\displaystyle e^{iz}=(\frac{1}{2})(\frac{\sqrt 3}{2}+ i \frac{1}{2})$

$\displaystyle e^{iz}=(\frac{1}{4})(\sqrt3+ i)$

I think that's right. It's just that when a lecturer has taken time out to write "this is a trick question" next to it, it implies it is much harder than it looks.

So is this the right answer?

Well it looks OK, but we can just do a numerical check:

Code:

>I
0+1i
>z=pi/6+I*log(2)
0.523599+0.693147i
>exp(I*z)
0.433013+0.25i
>
>
>(0.25)*(sqrt(3)+I)
0.433013+0.25i
>

RonL