# Proving for complex numbers

• Jun 8th 2013, 10:38 AM
yugimutoshung
Proving for complex numbers
How would you prove that z - 1/z = 2*j*sin(theta) ?
• Jun 8th 2013, 10:56 AM
Plato
Re: Proving for complex numbers
Quote:

Originally Posted by yugimutoshung
How would you prove that z - 1/z = 2*j*sin(theta) ?

If do you mean prove, then you must have more conditions on z.
Like $|z|=1$.
• Jun 8th 2013, 11:04 AM
yugimutoshung
Re: Proving for complex numbers
I think "z" refers to r*cis(theta)
• Jun 8th 2013, 11:32 AM
Plato
Re: Proving for complex numbers
Quote:

Originally Posted by yugimutoshung
I think "z" refers to r*cis(theta)

If $r\ne 1$ then $z-\frac{1}{z}\ne 2i\sin(\theta).$

So there is nothing to prove.
• Jun 8th 2013, 12:03 PM
MINOANMAN
Re: Proving for complex numbers
Quote:

Originally Posted by Plato
If $r\ne 1$ then $z-\frac{1}{z}\ne 2i\sin(\theta).$

So there is nothing to prove.

if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
for n=1 and r=1 .........obvious

check what you need.
• Jun 8th 2013, 12:23 PM
Plato
Re: Proving for complex numbers
Quote:

Originally Posted by MINOANMAN
if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
for n=1 and r=1 .........obvious

I don't understand if you meant comment for me or for the OP.

As I said it is true if $|z|=1$ it is true. Here is the standard proof.

Recall that if $z=\text{c}i\text{s}(\theta)$ then $\frac{1}{z}=\text{c}i\text{s}(-\theta)$.

Now it is just a exercise to finish.