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I actually stated that I didn't want to use this method on the course chat haha.What about:
$$\bar z^2 = (re^{-i\theta})^2 = r^2e^{-2i\theta} =r^2(\cos (-2\theta) +i\sin(-2\theta))$$
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.
I guess it's because $$\text{cis}(-\theta)=e^{-i\theta}$$.If z=reiθ, why does z¯=re−iθ?
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 $$\text{cis}(-\theta)=e^{-i\theta}$$.![]()