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- Thread starter Leo Liu
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If you try to justify that formula, where do you get stuck?

- #3

Leo Liu

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$$\bar z^2=r^2(\cos\theta-i\sin\theta)^2$$

$$\bar z^2=r^2(\cos(2\theta)-i\sin(2\theta))$$

$$\bar z^2=r^2(\cos2\theta+i\sin(-2\theta))$$

$$\bar z^2=r^2(\cos(-2\theta)+i\sin(-2\theta))$$

I got it. Thanks.

- #4

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$$\bar z^2 = (re^{-i\theta})^2 = r^2e^{-2i\theta} =r^2(\cos (-2\theta) +i\sin(-2\theta))$$

- #5

Leo Liu

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I actually stated that I didn't want to use this method on the course chat haha.

$$\bar z^2 = (re^{-i\theta})^2 = r^2e^{-2i\theta} =r^2(\cos (-2\theta) +i\sin(-2\theta))$$

Thank you though! It is very neat.

- #6

FactChecker

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You should study each part you doubt until it is intuitive to you. It is very fundamental and important.I actually stated that I didn't want to use this method on the course chat haha.

View attachment 293673

Thank you though! It is very neat.

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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- #9

Leo Liu

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I guess it's because $$\text{cis}(-\theta)=e^{-i\theta}$$.If z=reiθ, why does z¯=re−iθ?

- #10

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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)##.I guess it's because $$\text{cis}(-\theta)=e^{-i\theta}$$.

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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