Math Help - Complex Numbers

1. Complex Numbers

Write each of the given numbers in the form a+bi

a) (e^(9i)-e^(-9i))/(2i)

b) 2e^(9+((i*pi)/6))

I don't even know how to start...

(

2. On the a) part, use the fact that cosine is an even function and sine is an odd function.
Then $\begin{gathered} e^{9i} = \cos (9) + i\sin (9) \hfill \\
e^{ - 9i} = \cos ( - 9) + i\sin ( - 9) = \cos (9) - i\sin (9) \hfill \\
\end{gathered}$
.

3. Complex numbers - Euler's Formula

Hi -

You'll need to use Euler's Formula:

e^ix = cos(x) + i sin(x)

Then use the fact that cos(-x) = cos(x) and sin(-x) = -sin(x) to simplify the result.

It's also useful to note that e^(p + qi) = e^p . e^qi.

Is that any help?

4. Thanks!! =)

One more question,
I just don't know what they're asking and why two of the data points have two numbers and why the last one has three...
Find a (complex) quadratic polynomial P(z) = a+bz+cz^2 which interpolates the data (-1,i), (0,i), and (5,i,0)

a=
b=
c=