1
$\begingroup$

Could anyone help me separate this into partial fractions: $$\frac{\cos(z)}{z^2+1}$$ where $z=x+iy$. I've factored the denominator to get $$\frac{\cos(z)}{(z+i)(z-i)}$$ but I'm not really sure where to go from there. Thanks

$\endgroup$
1
$\begingroup$

You want to do it without the cosine? $$ \frac{1}{(z+i)(z-i)} = \frac{A}{z+i}+\frac{B}{z-i} $$ where $A,B$ are complex constants. Solve for $A,B$ as in the real case. (But use complex arithmetic.)

.........
Then you can multiply by $\cos(z)$, it does not vanish at $i$ or at $-i$.

$\endgroup$
  • $\begingroup$ So the answer is $$\frac{icos(z)}{2(z+i)} - \frac{icos(z)}{2(z-i)}$$? $\endgroup$ – Arron Apr 4 '15 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.