1. ## Complex numbers.

The question is:

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

(careful! This is a trick question.)

What I did:

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

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

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

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

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

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

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?

2. Originally Posted by Showcase_22
The question is:

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

(careful! This is a trick question.)

What I did:

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

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

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

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

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

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

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 let's see...

If $z = x + iy$ then $e^z = e^x \ {\rm{cis}} (y)$.

3. ahhhh! The last bit you wrote ended up with a Latex error!

4. Originally Posted by Showcase_22
ahhhh! The last bit you wrote ended up with a Latex error!
Corrected, but what it says you knew already!

RonL

5. Originally Posted by Showcase_22
The question is:

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

(careful! This is a trick question.)

What I did:

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

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

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

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

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

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

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

6. Corrected, but what it says you knew already!
Actually, i'm not sure I do :s

What does $e^z=e^x cis (y)$ mean? More specifically, what does "cis" stand for?

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

>exp(I*z)
0.433013+0.25i
How did you work out that bit? Can your calculator work out complex numbers?

7. Originally Posted by Showcase_22
Actually, i'm not sure I do :s

What does $e^z=e^x cis (y)$ mean? More specifically, what does "cis" stand for?
$cis(y)$ is a shorthand way of writing $\cos{y} + i\sin{y}$, which is a complex number of magnitude 1 written in polar co-ordinates.

By definition of $z = x + iy$, we have

$e^z = e^{x+iy} = e^x e^{iy} = e^x cis(y)$

because, by Euler's formula

$e^{i\theta} = cis(\theta)$

8. Originally Posted by Showcase_22
Actually, i'm not sure I do :s

What does $e^z=e^x cis (y)$ mean? More specifically, what does "cis" stand for?
It actualy means very little, it is a relativly recently invented notation for:

$\cos(x)+{\bold{i}}\sin(x)$

but such a notation is unneccessary, and redundant.

How did you work out that bit? Can your calculator work out complex numbers?
Well mine can, yes

RonL

9. oh right!

That cis notation looks redundant but it is shorter than writing $cos x+i sinx$. I might start using it on my whiteboard since I always run out of space on it!

My calculator can't do complex numbers. Well I don't think it can!

Does a TI-84 have a complex number setting? I haven't really used it and I didn't take the manual with me to university.

Also, for a trick question, that wasn't that bad. I guess lecturers write that there to freak students out!