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I'm having a lot of problem with the first term. If I try to rationalize the fraction, I get an expression too long to work with. Similarly, substituting from the second term gives horrible results.

Breaking the modulus apart and squaring gives me an incorrect result, and I just don't know how to approach the problem anymore. I'm probably making a dumb mistake along the lines, but I'd be grateful if someone could solve this step by step.

EDIT: Wrote the thing incorrectly, switched the denominator and enumerator. Sorry for anyone who solved the wrong thing, but they should be similar enough for a simple switch, no?

Michael Hardy
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John Doe
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2 Answers2

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If $z=a+ib$ then $a^{2}+b^{2}=5$ is the second equality, and the first equality is $$a^{2}-2a+1+b^{2}=2(a^{2}+b^{2}-2b+1),$$ which is equivalent to $a^{2}+b^{2}-4b+2a+1=0$. Substitute the first to the second to get $-4b+2a+6=0$. Can you take it from here?

T. Eskin
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$$\left| \frac{z-1}{z-i}\right|^2 = 1/2$$

$$\implies \frac{z-1}{z-i}\frac{\bar{z}-1}{\bar{z}+i}=1/2$$

$$\implies \frac{|z|^2-(z+\bar{z})+1}{|z|^2+i(z-\bar{z})+1}=1/2$$

$$\implies \frac{5-2\Im(z)+1}{5-2\Re (z)+1}=2$$

$$\implies 2(5-2\Re (z)+1)=5-2\Im(z)+1$$

$$6-4\Re (z)+2\Im(z)=0$$

$$\implies \Re(z)=3/2\text{ and } \Im(z)=0$$

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