Finding z-(1/z) Given z=r(cosx+isinx)

[araz1]

Hi
Show that z-(1/z)=i2rsinx, given z=r(cosx+isinx)
Thanks.

 

[I like Serena]

That is not the case.
Did you perhaps mean $z-\bar z$?
The notation $\bar z$ is the so called complex conjugate of $z$, which is $\bar z = r(\cos x-i\sin x)$.

 

[araz1]

No it is not z minus its conjugate. it is z minus (1 over z).

 

[I like Serena]

No it is not z minus its conjugate. it is z minus (1 over z).

Then it is not the case.
Consider for instance $z=2$. Then we get $2-\frac 12=\frac 32\ne i\cdot 2\cdot \sin 0=0$.

 

[araz1]

Your example makes sense, but that is a question in a book. Does it matter that z is in polar form?

 

[I like Serena]

No Polar form is just another way to write a complex number. It does not affect calculations.
Likely there is a typo in the book then.

 

[araz1]

Yes there must be an error. Thank you for that.