Complex conjugate of a pole is a pole?

[docnet]

Homework Statement

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Relevant Equations

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This isn't a homework problem, but a more general question.

Let $f$ be a function with two singular points $r$ and its complex conjugate $r^*$.

let
$$f=\frac{g}{z-r} \quad \text{and assume} \quad g(r)\neq 0$$
so $r$ is a simple pole of $f$.

we have conjugates that are singular points of $f$,
can we say that $r^*$ is also a simple pole of $f$ because it is a complex conjugate of $r$? if so, what property of complex functions can we invoke to avoid doing the same calculations for $r^*$?

 

[hutchphd]

Why are you defining u?? But OK
The function $$f(z)=\frac 1 {z^2-a^2} $$ has poles at $$z=\pm a $$ This is a characteristic of the function and is not generally true.

 

[Office_Shredder]

. -r is not the complex conjugate of r in most cases.

And I think $f=\frac{1}{(z-1)(z+1)^2}$ satisfies your post but does not have a simple pole at both roots.

 

[FactChecker]

can we say $-r$ is also a simple pole of $f$ because it is a complex conjugate of $r$?

$-r$ is not the complex conjugate of $r$ unless $r$ is purely imaginary.

if so, what property of complex functions can we invoke to avoid doing the same calculations for $-r$?

If $g(z)$ is a polynomial with real coefficients and $r$ is a root, then $\bar {r}$ is also a root.

PS. I assume that you are using '$r$' to designate a root rather than a real

 

[Office_Shredder]

No. If you insist -r is the complex conjugate of r then r is imaginary, so let's say $r=i$. Then consider
$f(z)=\frac{1}{(z-i)(z+i)^2}$

 

[docnet]

that makes sense. sorry, think i was confusing this with the case of real valued functions always having complex valued roots in conjugate pairs. my brain is not working its best tdaoy