Power of a complex conjugate

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  • #1
Leo Liu
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Can someone please tell me why this is true? This isn't exactly the De Moivre's theorem. Thank you.
 

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  • #2
PeroK
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If you try to justify that formula, where do you get stuck?
 
  • #3
Leo Liu
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$$\bar z^2=(x-yi)^2$$
$$\bar z^2=r^2(\cos\theta-i\sin\theta)^2$$
$$\bar z^2=r^2(\cos(2\theta)-i\sin(2\theta))$$
$$\bar z^2=r^2(\cos2\theta+i\sin(-2\theta))$$
$$\bar z^2=r^2(\cos(-2\theta)+i\sin(-2\theta))$$
I got it. Thanks.
 
  • #4
PeroK
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What about:
$$\bar z^2 = (re^{-i\theta})^2 = r^2e^{-2i\theta} =r^2(\cos (-2\theta) +i\sin(-2\theta))$$
 
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  • #5
Leo Liu
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What about:
$$\bar z^2 = (re^{-i\theta})^2 = r^2e^{-2i\theta} =r^2(\cos (-2\theta) +i\sin(-2\theta))$$
I actually stated that I didn't want to use this method on the course chat haha.
1638798004769.png

Thank you though! It is very neat.
 
  • #6
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I actually stated that I didn't want to use this method on the course chat haha.
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Thank you though! It is very neat.
You should study each part you doubt until it is intuitive to you. It is very fundamental and important.
In words explain why:
If ##z=re^{i\theta}##, why does ##\bar{z}=re^{-i\theta}##?
Then why does ##\bar{z}^2=re^{-i2\theta}##?
etc.
 
  • #7
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Many mathematicians consider Euler's Formula, ##e^{i\theta} = \cos{\theta} + i\sin{\theta}##, to be the most important equation in mathematics. You should get very comfortable with using it.
 
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  • #8
PAllen
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You can also just to algebra on the conjugate expressed in terms of ##\theta##, and then apply double angle trig formulas, if you want to see it verified brute force. It works out very straightforward this way, though, of course, using exponentials is far more elegant.
 
  • #9
Leo Liu
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If z=reiθ, why does z¯=re−iθ?
I guess it's because $$\text{cis}(-\theta)=e^{-i\theta}$$. :wink:
 
  • #10
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I guess it's because $$\text{cis}(-\theta)=e^{-i\theta}$$. :wink:
Yes. Negating the imaginary part of ##z=x+iy = r(\cos\theta+ i\sin\theta)## to get ##\bar{z}= x-iy = r(\cos\theta- i\sin\theta)## is the same as negating the argument, ##\theta##, to get ##\bar{z}=r(\cos{(-\theta)} + i\sin{(-\theta)})= r(\cos\theta- i\sin\theta)##.

It looks like every line in your original post is correct (I didn't look hard at it.), but it is not clear how you got some lines and if you understood it. At least in the beginning, it is good practice to really spell everything out. You can skip some details after you are well beyond that level, but not before.
 
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