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Let $\alpha \neq 1$ be a complex number such that the distance from $\alpha^2$ to 1 is twice the distance from $\alpha$ to 1, while the distance from $\alpha^4$ to 1 is four times the distance from $\alpha$ to 1. Enter all possible values of $\alpha,$ separated by commas.

I have no idea how to do this. Can someone help me?

sumi
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  • A good place to start might be by determining how to compute the distance from an arbitrary complex number to zero. You might start by thinking about this computation in components (either in terms of the real and imaginary components; or in terms of the modulus and phase, i.e. in polar coordinates). – Xander Henderson Apr 05 '19 at 01:26

2 Answers2

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You have $$ |\alpha^4-1|=2|\alpha^2-1|=4|\alpha-1|. $$ Let's use bars to denote complex conjugate. Actually we do not need to write $\alpha$ as $a+bi$. We have $$ |\alpha^4-1|^2=(\alpha^4-1)(\bar\alpha^4-1)=(\alpha^2+1)(\bar\alpha^2+1)(\alpha^2-1)(\bar\alpha^2-1)=|\alpha^2+1|^2|\alpha^2-1|^2,\\ |\alpha^4-1|=|\alpha^2+1||\alpha^2-1| $$ So, $|\alpha^2+1|=2$. Similarly, $|\alpha+1|=2$

We can rewrite those results as $$ \alpha^2\bar\alpha^2+\alpha^2+\bar\alpha^2=3\\ \alpha\bar\alpha+\alpha+\bar\alpha=3 $$ From the second equation we get $$ \bar\alpha=\frac{1-\alpha}{1+\alpha}. $$ Substituting into the first equation, and factorize a bit, we get $$ (1+\alpha^2)(3-\alpha)^2-(3-\alpha^2)(1+\alpha)^2=0 $$ This gives us the trivial solution $\alpha=1$. To find other solutions, we can simplify the above equation. We can then obtain $$ (\alpha-1)(\alpha^3-\alpha^2+3\alpha-3)=(\alpha-1)(\alpha^2+3)(\alpha-1)=0 $$ The other two sultions, $$ \alpha=-\sqrt{3} i,+\sqrt{3} i $$ Satisfy the requirements of the question.

Ma Joad
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An alternative method:

Use the first condition to find$$|\alpha^2-1|=2|\alpha-1|\\|\alpha+1|=2\\\alpha=-1+2e^{i\theta}$$

Use the second condition to get $$|\alpha^4-1|=4|\alpha-1|\\|\alpha^3+\alpha^2+\alpha+1|=4\\|(2e^{i\theta}-1)^3+(2e^{i\theta}-1)^2+2e^{i\theta}|=4\\\left|8e^{3i\theta}-12e^{2i\theta}+6e^{i\theta}-1+4e^{2i\theta}-4e^{i\theta}+1+2e^{i\theta}\right|=4\\\left|2e^{2i\theta}-2e^{i\theta}+1\right|=1\\2e^{2i\theta}-2e^{i\theta}+1=e^{i\phi}\\2\cos2\theta-2\cos\theta+1=\cos\phi\in[-1,1]\\2\sin2\theta-2\sin\theta=0$$This second equation tells us $$(4\cos\theta-2)\sin\theta=0$$So $\theta=\pi,\pm\frac\pi3$. If $\theta=\pi$, then the first equation gives $5=\cos\phi$ which is not solvable for real $\phi$, so this is not a solution. Therefore have $\theta=\pm\frac\pi3$.

$$\alpha=-1+2\left(\frac12\pm i\frac{\sqrt3}2\right)=\pm\sqrt3i$$

John Doe
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