real vs imaginary

Jan 2014
40
0
Hope I have the right forum section for this question.

For a complex variable z=x+iy i = -1^1/2
What are the real and imaginary parts of g(z)=e^sin(z)

This question confuses me, since i is a negative radical it is imaginary therefore this would make z imaginairy correct? the rest of e^sin(z) would be considered real?
 

v8archie

Math Team
Dec 2013
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2,682
Colombia
\(\displaystyle 2i\sin{(z)} = e^{iz} - e^{-iz} = e^{i(x + iy)} - e^{-i(x + iy)} = e^{(-y + ix)} - e^{(y - ix)} = e^{-y}e^{ix} - e^{y}e^{-ix}\)
and
\(\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}\)

Does that help at all?
 

v8archie

Math Team
Dec 2013
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2,682
Colombia
Oh, and z = x + iy is complex because it has both real (x) and imaginary (iy) parts.
 
Jan 2014
40
0
v8archie said:
\(\displaystyle 2i\sin{(z)} = e^{iz} - e^{-iz} = e^{i(x + iy)} - e^{-i(x + iy)} = e^{(-y + ix)} - e^{(y - ix)} = e^{-y}e^{ix} - e^{y}e^{-ix}\)
and
\(\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}\)

Does that help at all?
I do not follow, were did 2isin at the start come from?
 

v8archie

Math Team
Dec 2013
7,712
2,682
Colombia
\(\displaystyle \sin{z) = \frac{e^{iz} - e^{-iz}}{2i}\)
It''s a standard identity that can be derived from
\(\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}\)
by setting \(\displaystyle \theta = z\) to get one equation and \(\displaystyle \theta = -z\) to get a second, and then subtracting one from the other.