Question $7(1$ point$)$

Let $Cbe$ the circle around the origin $of$ radius $0\mathrm{.}5,$ traversed once $in$ the positive

direction. Compute

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{arctan(3\mathrm{.}6z^2)}{z^7}dz$

The answer $is$ a purely imaginary number, enter its imaginary part rounded $to$ two

decimals. Hint: You can avoid a lot $of$ work here $by$ using a power series expansion $of$

the function $in$ the integral $(using$ the Maclaurin series $ofarctan$ derived $in$ Notes $24),$

and then using $e.g.$ the Cauchy Integral Formula.

Bonus Exercise: Why could $we$ not use the unit circle $as$ a contour here?

Your Answer:

Answer

Question $8(1$ point$)$

$In$ a punctured neighbourhood $ofi,$ develop $f(z)=\frac{1}{1+z^2}\frac{1}{z+i}$ into a geometric series $in$ powers $ofw$ :$=$

$z-iasin$ Notes $25,$ a possibly faster alternative would $beto$ compute its Taylor Series

around i with Taylor's "higher derivatives" formula for the coefficients from Calculus

$II.-\frac{1}{2}(z-i{)}^{-1}+\frac{i}{4}+\frac{1}{8}(z-i)+-$dots and converges

exactly $on{B}_2\frac{i}{\frac{?}{?}}\{i\}$

$...$ starts with $-\frac{i}{2}(z-i{)}^{-1}+\frac{1}{4}+\frac{i}{8}(z-i)+-$dots and converges

exactly $on{B}_2\frac{i}{\frac{?}{?}}\{i\}$

$...$ starts with $\frac{i}{2}(z-i{)}^{-1}-\frac{1}{4}-\frac{i}{8}(z-i)+-$dots and converges exactly

$on{B}_2\frac{i}{\frac{?}{?}}\{i\}$

$...$ starts with $-\frac{i}{2}(z-i{)}^{-1}+\frac{1}{4}+\frac{i}{8}(z-i)+-$dots and converges

exactly $on{B}_1\frac{i}{\frac{?}{?}}\{i\}$