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Find all complex numbers $z$ such that $|z|=|\sqrt{2} +z|=1$. Prove that each of these satisfies $z^8=1$ I'm a little unsure of where to start. I know that the modulus of $z$ is the distance from the origin. The book states

$|z|=1$ implies $z=\cos\theta + i\sin\theta$ for some $\theta$

This part makes sense. But the part I don't understand is this:

$|\sqrt{2} +z|=1$ implies $(\cos\theta + \sqrt{2})^2 +\sin^2\theta = 1$

Any help would be appreciated.

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    hint: Find the intersection points of two circles. – 2017-02-21
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    To be clear, this $\sqrt 2$, not $\sqrt{-2}$, right? – 2017-02-21

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The intersection occurs at $z=x\pm iy$, where $x=-\frac12\sqrt2$ and $x^2+y^2=1\implies y=\frac12\sqrt2$

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Starting from $|\sqrt{2} + z| = 1$, square both sides and re-write $z$ in the way you've given: $$|\sqrt{2} + \cos\theta + i \sin\theta|^2 = 1$$

So we've separated out the real and imaginary parts, and we now can say that $1=|\sqrt{2} + \cos\theta + i \sin\theta|^2 = (\sqrt{2} + \cos\theta)^2 + \sin^2\theta$. Now expand the brackets and use the (hopefully) familiar trig identities to finish off the solution.

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    We are grouping $\sqrt{2}$ with $cos\theta$ because of it being real? – 2017-02-21
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    @Lanous Yes, exactly – 2017-02-21
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$$|z|=|\sqrt{2} +z|=1 \Rightarrow \\ |z|^2=|\sqrt{2} +z|^2 \Rightarrow \\ z \bar{z}= z \bar{z}+(z+\bar{z}) \sqrt{2}+2 \Rightarrow \\ z+\bar{z}=\sqrt{2}$$

Now, since $$z+\bar{z}=\sqrt{2} \\ z \cdot \bar{z}=1$$ it follows that $z, \bar{z}$ are the two roots of $$z^2-z \sqrt{2}+1=0$$