1

So I encountered this one question today in my math book, and I don't know how to get the right answer even though it seems really easy, I just wanna know how to do it so i can get some sleep.

This is the question:

When the complex number $z=r(\cos\theta + i \sin\theta)$ show that $z-\frac{1}{z}=i(2r\sin\theta)$

I haven't done much for this question other than do this using the actual complex number: $$\frac{z^2-1}{z}$$

Can anyone show me how to do prove this question, any help is appreciated.

Bernard
  • 175,478
mikejacob
  • 327

2 Answers2

1

The identity is not true, indeed we have that

$$z-\frac1z=z-\frac{\bar z}{|z|^2}=r(\cos\theta + i \sin\theta)-\frac1r(\cos\theta - i \sin\theta)=\left(r-\frac1r\right)\cos \theta+\left(r+\frac1r\right)i\sin \theta$$

the identity is true only for $\left(r-\frac1r\right)=0$ and $\left(r+\frac1r\right)=2r$ that is $r=1$ or $z=e^{i\theta}$.

user
  • 154,566
  • Are you sure that the answer in the book is wrong? If so, thank you so much for this, you've saved me a lot of time! – mikejacob Sep 30 '20 at 21:05
  • @mikejacob From we can see I can't give other explanations. This is true only for $r=1$. Is there some assumption for $z$? – user Sep 30 '20 at 21:09
  • No that's all of the information they gave, so i guess we would have to assume that $r=1$. Thanks again – mikejacob Sep 30 '20 at 21:22
  • @mikejacob I really can't see other explanations. – user Sep 30 '20 at 21:23
  • Is it a cheap edition? With some very bad type setting, $\bar z$ may have been mistaken for $\frac{1}{z}$. I saw a cheap Asian reprint once with a table which said $100 = 1, 101 = 10, 102 = 100, 103 = 1000$. It took me a little while to figure out what it should have said. – badjohn Sep 30 '20 at 22:09
  • @badjohn That's also could be an explanation! – user Sep 30 '20 at 22:11
1

What's equal to $\:2ir\sin\theta\:$ is $\:z -\color{red}{\bar z}$.

Now $\bar z=\dfrac 1 z\:$ if and only if $z$ has modulus $1$.

Bernard
  • 175,478