Finding Partial Fractions and Integrating over Unit Circle for Complex Integral?

[John O' Meara]

write f(z) in terms of partial fractions and integrate it counterclockwise over the unit circle where $$f(z)=\frac{2z+3\iota}{z^2+\frac{1}{4}} \\$$
$$2\int\frac{z}{z^2 + \frac{1}{4}} dz \ + \ 3\iota \int\frac{1}{z^2 + \frac{1}{4}}dz \\ \ \mbox{. Also }$$
$$\ z(t) = \exp{\iota t} (0 \leq \ t \ \leq2\pi) \ \frac{dz}{dt}=\iota \exp{\iota t} \\ \ \mbox{therefore} \ f(z(t))=\frac{\exp{\iota t}}{\exp{2\iota t} + \frac{1}{4}}\\ \$$
$$\ 2\iota \int_0 ^{2\pi} \frac{\exp{2\iota t}}{\exp{2\iota t} + \frac{1}{4}}dt \\$$ I let $$u=\exp{2\iota t} + \frac{1}{4} \\ \ \mbox{ therefore I get } \ \\ \ \int_1 ^1 \frac{1}{u} du=0\\$$. I also get 0 for the 2nd integral expression. Please check my results for error. Thanks.

 

[mighty2000]

Your approach and steps are correct. However, there is a small error in your final result. When substituting u=exp(2it)+1/4, the limits of integration should change as well. The correct limits should be from u=1 to u=1+1/4=5/4. Therefore, the final result should be:

2i∫0^2π (1/5) exp(2it)/(exp(2it)+1/4) dt = (2i/5)∫1^5/4 (1/u) du = (2i/5) ln(5/4) = (2i/5) ln(5) - (2i/5) ln(4)

So, the final result for the integral over the unit circle is:

f(z) = (2i/5) ln(5) - (2i/5) ln(4)

Please note that this result may still be simplified further, but the steps and approach you have taken are correct.