residue

**alexandrabel90** · May 23rd 2011, 11:55 AM

how to find the residue of -2/ (3z^2 - 10iz -3) at z=i/3.

i know that this has poles at i/3 and 3i

so i used the formula of

-2 /( i/3- 3i) and got 3i/4 where the denominator is (z_ -3i)

but when i used

-2 /( 6z_ - 10i) and sub z_=i/3, i got the answer as -i/4. (the denominator here is done by differentiation)

i thought that both methods should give the same answer?

**TheEmptySet** · May 23rd 2011, 12:17 PM

Originally Posted by alexandrabel90

how to find the residue of -2/ (3z^2 - 10iz -3) at z=i/3.

i know that this has poles at i/3 and 3i

so i used the formula of

-2 /( i/3- 3i) and got 3i/4 where the denominator is (z_ -3i)

but when i used

-2 /( 6z_ - 10i) and sub z_=i/3, i got the answer as -i/4. (the denominator here is done by differentiation)

i thought that both methods should give the same answer?

For simple poles we have the formula

$\text{Res}(f,z_0)=\lim_{z \to z_0}(z-z_0)f(z)$

So if we factor the function we get

$f(z)=\frac{-2}{(3z-i)((z-3i))}$

$\text{Res}\left(f,\frac{i}{3} \right)=\lim_{z \to \frac{i}{3}}\left(z-\frac{i}{3} \right)\frac{-2}{3z^2-10iz-3}=\lim_{z \to \frac{i}{3}}\frac{-2}{3(z-3i)}=-\frac{i}{4}$

Always be careful with formula's.

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