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...

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

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=

5. ## Complex numbers - quadratic

Hi -

I'm not sure what they're asking either. Perhaps we want a quadratic for y in terms of x: y = a + b.x + c.x^2, given (x, y) pairs as follows:

(-1, i), (0, i), and (5 + i, 0)

This does make sense. If this is what is wanted, just substitute for x and y for each of the three pairs of values, and solve the three equations for a, b and c.

In any event, I really don't know what (5,i,0) means unless we have a point in three dimensions whose coordinates are (x, y, z). (But then why are the other two only given as (x, y) pairs?)