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

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Discussion Overview

The discussion revolves around the expression z - (1/z) given z = r(cos x + i sin x). Participants are examining whether the expression can be shown to equal i2rsin x, and they are considering the implications of using polar form for z.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant requests a demonstration that z - (1/z) equals i2rsin x.
  • Another participant suggests that the original request may have been intended to refer to z - z̅ (the complex conjugate), noting the definition of the conjugate.
  • A different participant clarifies that the expression in question is indeed z minus (1/z), not the conjugate.
  • One participant provides a counterexample using z = 2, showing that the expression does not yield the expected result, indicating a potential issue with the original claim.
  • Another participant questions the relevance of the polar form of z in the calculations.
  • One participant asserts that polar form does not affect calculations and suggests a possible typo in the book from which the problem originates.
  • A later reply concurs that there must be an error in the book, thanking the previous participant for their input.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the original expression and whether it can be shown as claimed. There is no consensus on the correctness of the original statement, and multiple views on the interpretation of the expression are present.

Contextual Notes

Participants note that the discussion may be influenced by the specific form of z and the potential for typographical errors in the source material.

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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