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

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The discussion revolves around the expression z - (1/z) given z in polar form as r(cos x + i sin x). Participants clarify that the expression does not equate to i2rsinx, as demonstrated with a counterexample using z = 2. The conversation highlights that the polar form of z does not influence the calculations involved. Ultimately, the consensus is that there is likely a typo in the book prompting the question. The discussion concludes with an acknowledgment of the potential error in the original problem.
araz1
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Hi
Show that z-(1/z)=i2rsinx, given z=r(cosx+isinx)
Thanks.
 
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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 $\bar z = r(\cos x-i\sin x)$.
 
No it is not z minus its conjugate. it is z minus (1 over z).
 
araz said:
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 $2-\frac 12=\frac 32\ne i\cdot 2\cdot \sin 0=0$.
 
Your example makes sense, but that is a question in a book. Does it matter that z is in polar form?
 
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.
 
Yes there must be an error. Thank you for that.
 

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