Proving for complex numbers

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

Re: Proving for complex numbers

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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 .

Re: Proving for complex numbers

I think "z" refers to r*cis(theta)

Re: Proving for complex numbers

Quote:

Originally Posted by

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

If then

So there is nothing to prove.

Re: Proving for complex numbers

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Originally Posted by

**Plato** If

then

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.

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 it is true. Here is the standard proof.

Recall that if then .

Now it is just a exercise to finish.