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...
thanks in advance!!
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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} $.
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?
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)
Write your answers in Cartesian form x=iy
a=
b=
c=
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?)