Complex numbers proof

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Kiev001

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Not sure how to approach this question:

Z = r(cos(x) + isin(x))
Show that:
Z - 1/Z = i(2rsin(x))

I've tried substituting in and rearranging and also converting to exponential form first, even using de moivres theorem but I can't seem to get to their answer. Any help would be appreciated!
 
I don't believe this is true.


If \(\displaystyle r=1\)

[MATH]z - \dfrac 1 z = \\ e^{i\theta}-e^{-i\theta} = 2i \sin(\theta)\\ \text{but when you throw $r \neq 1$ into the mix you get $r$ multiplying one term and $r^{-1}$ the other so it doesn't work out} [/MATH]
 
Thank you for taking the time to reply! I will try to find the question again to make sure I didn't make a mistake when copying it down.
Also maybe Im missing something but is there an identity that shows that the sum of those exponentials being equal to sine?
 
Kiev001 said:
Z = r(cos(x) + isin(x))
Show that:
Z - 1/Z = i(2rsin(x))
Click to expand...
Here is a most useful identity: \(\displaystyle z^{-1}=\frac{\overline{~z~}}{|z|^2}\).
Moreover, if \(z=\cis(x)\) then \(|z|=1\) thus \(z-z^{-1}=z-\overline{z}=2i\Im(z)=2i\sin(x)\)
 
pka said:
Here is a most useful identity: \(\displaystyle z^{-1}=\frac{\overline{~z~}}{|z|^2}\).
Moreover, if thus \(z-z^{-1}=z-\overline{z}= 2i\Im(z)=2i\sin(x)\)
Click to expand...
Thank you for you reply! That mostly makes sense, is there a way to show that when \(z=\cis(x)\) then \(|z|=1\) , or is that a given? Also, what does the notation after the '2i' represent here: \(2i\Im(z)\) ?
 
Romsek said:
I don't believe this is true.


If \(\displaystyle r=1\)

[MATH]z - \dfrac 1 z = \\ e^{i\theta}-e^{-i\theta} = 2i \sin(\theta)\\ \text{but when you throw $r \neq 1$ into the mix you get $r$ multiplying one term and $r^{-1}$ the other so it doesn't work out} [/MATH]
Click to expand...
Sorry, to make it clearer from previous comment, is the following an identity or is this something that can be worked out?
[MATH] e^{i\theta}-e^{-i\theta} = 2i \sin(\theta)\\ [/MATH]Sorry, I'm just not familiar with this!
 
Kiev001 said:
Thank you for you reply! That mostly makes sense, is there a way to show that when \(z=\cis(x)\) then \(|z|=1\) , or is that a given? Also, what does the notation after the '2i' represent here: \(2i\Im(z)\) ?
Click to expand...
The short answer is \(\cos^2(x)+\sin^2(x)=1\) Why is that relevant?
If \(z=a+bi\) then \(\Re(z)=a~\&~\Im(z)=b\) using that notation \(|z|=\sqrt{\Re^2(z)+\Im^2(z)}=\sqrt{a^2+b^2}\).
Thus if \(z=\cis(x)=\cos(x)+i\sin(x)\) so \(|\cos(x)+i\sin(x)|=\sqrt{\cos^2(x)+sin^2(x)}=1\)
 
pka said:
The short answer is \(\cos^2(x)+\sin^2(x)=1\) Why is that relevant?
If \(z=a+bi\) then \(\Re(z)=a~\&~\Im(z)=b\) using that notation \(|z|=\sqrt{\Re^2(z)+\Im^2(z)}=\sqrt{a^2+b^2}\).
Thus if \(z=\cis(x)=\cos(x)+i\sin(x)\) so \(|\cos(x)+i\sin(x)|=\sqrt{\cos^2(x)+sin^2(x)}=1\)
Click to expand...
Thank you very much! Really appreciate you taking your time to reply and help out further!
 
Kiev001 said:
Sorry, to make it clearer from previous comment, is the following an identity or is this something that can be worked out?
[MATH] e^{i\theta}-e^{-i\theta} = 2i \sin(\theta)\\ [/MATH]Sorry, I'm just not familiar with this!
Click to expand...

It can be derived a number of ways.

[MATH]\cos(\theta)=\dfrac{e^{i\theta}+e^{-i\theta}}{2},~\sin(\theta)=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}[/MATH]
[MATH]\text{We know that $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ from Euler's formula.}\\ \text{and $e^{-i\theta} = \cos(\theta) - i \sin(\theta)$ thus}\\ \text{$e^{i\theta}+e^{-i\theta}=2\cos(\theta)$ and $e^{i\theta}-e^{-i\theta} = 2i\sin(\theta)$}[/MATH]
 
I appreciate everyone's responses and reminders. I just realised that I knew z could be written as cis(x) or as the exponential, but didn't put two and two together in my head to realise that cis(x) could therefore be written as equal to the exponential which also confused me a bit. Thanks again for all the help!
 
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