where c is the positively oriented circle So the formula is So my answer is But according to my teacher it is over 27
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f(z) is not e^z in this question \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}
Originally Posted by adam_leeds Sorry still dont get it.
As you can see, only one of the poles (z=-3) lies within the given circle, so we let and use Cauchy's integral formula for f(-3):
we have two singular points -3,3 but 3 is outside the circle but -3 is inside so the singular point for the function is -3 the function is
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