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Let set $$C=\{z\in \mathbb{C}:\sqrt{2}|z|=(i-1)z\}$$ I think C is empty , because you could put it in this way$$|z|=-\frac{(1-i)z}{\sqrt{2}}$$ but would like a second opinion.

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The set condition is the same as $2z\bar{z}=(i-1)^2z^2$. If you let $z=u+iv$ then $\bar{z}=u-iv$, where $\bar{z}$ is the conjugate of $z$. Expand and solve. – poirot Jan 11 '13 at 11:06
up vote 4 down vote accepted

I don't understand your argument, but in any case the result is wrong.

The left-hand side is the absolute value of the right-hand side. A complex number is equal to its absolute value if and only if it's a non-negative real number. Thus $(i-1)z$ must be a non-negative real number, so $z$ must be of the form $(i+1)\lambda$ with $0\ge\lambda\in\mathbb R$.

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so would it be a line? @joriki – Mykolas Jan 11 '13 at 10:26
@Mykolas: No. Try drawing it. Multiplication by $i-1$ is an isometry, so it must have the same shape as the set of non-negative real numbers. – joriki Jan 11 '13 at 10:27
so it's a ray from the origin? @joriki – Mykolas Jan 12 '13 at 9:26
@Mykolas: Yes. $ $ – joriki Jan 13 '13 at 0:45

$$C=\{z\in \mathbb{C}:\sqrt{2}|z|=(i-1)z\}$$ let $z=a+bi,a,b\in\mathbb R$ then $$\sqrt{2}|a+bi|=(i-1)(a+bi)$$ $$\sqrt{2}|a+bi|=ai+bi^2-a-bi$$ $$\sqrt{2}|a+bi|=ai-b-a-bi$$ $$\sqrt{2}|a+bi|=-b-a+(a-b)i$$ follow that $a-b=0$ and $\sqrt{2}|a+bi|=-b-a$ or $a=b$ and $\sqrt{2}|a+ai|=-2a\geq 0$ so $a=b\leq 0$ or $z=a+ai=a(1+i)$ and finally $$C=\{z=a(1+i),a\leq 0\}\neq\emptyset$$

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Hint: Set $z = Re^{i \theta}$. What can you say about $\theta$?

Alternative approach hint: Multiplication by $(i-1)$ is equivalent to what geometric transformation?

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