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If $|z^2-1|=|z|^2+1$, how do we show that $z$ lies on imaginary axis ?

I understand that I can easily do this if I substitute $z=a+ib$. How do we solve it using algebra of complex numbers without the above substitution ?

My Attempt: $$ |z|^2+|1|^2=|z-1|^2+2\mathcal{Re}(z)=|z^2-1|\\ 2\mathcal{Re}(z)=|z^2-1|-|z-1|^2=|(z+1)(z-1)|-|z-1|.|z-1|\\=|z+1|.|z-1|-|z-1|.|z-1| $$ How do I proceed further and prove $\mathcal{Re}(z)=0$ ?

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  • $\begingroup$ What do you mean by "avoiding the substitution"? There is no algebraic way to introduce $|\cdot|$ in complex numbers, so there's no way you could avoid it. I would say that considering either $\mathcal{Re} (z)$ or $\bar{z}$ counts as using "the substitution" (although the latter is more elegant). $\endgroup$ – Michał Miśkiewicz Dec 30 '16 at 17:58
  • $\begingroup$ @MichałMiśkiewicz I understand that. To prove the equalities applied above we need the substitution. Just curious to find whether i can prove it without directly substituting $z=a+ib$. $\endgroup$ – ss1729 Dec 30 '16 at 18:03
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    $\begingroup$ I just wanted to point out that the problem you stated is impossible (algebra of complex numbers is not sufficient even for stating the problem) and that the one you had in mind is not of mathematical nature (rather aesthetic). Anyway, I guess Nathan's answer is probably more or less what you seek. $\endgroup$ – Michał Miśkiewicz Dec 30 '16 at 18:11
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Use the fact that $|z|^2 = z\bar{z}$.

Squaring both sides of the given equality yields \begin{align} |z^2-1|^2 &= (z\bar{z} + 1)^2\\ (z^2 - 1)(\bar{z}^2 - 1) &= (z\bar{z} + 1)(z\bar{z}+1)\\ z^2 + 2z\bar{z} + \bar{z}^2 &= 0\\ (z + \bar{z})^2 &= 0\\ z = -\bar{z} \end{align} from which it follows that the real part of $z$ is $0$. (I skipped some simple algebra steps above.)

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  • $\begingroup$ Thanx...but could u comment on my attempt, how do i finish the proof that way ? $\endgroup$ – ss1729 Dec 30 '16 at 17:57
  • $\begingroup$ I'm not sure how to finish it your way (or even if you can), since you can't really get rid of the absolute values, or do anything with them, without using $|z|^2 = z\bar{z}$ or splitting $z$ up into $z=a+ib$. $\endgroup$ – Nathan H. Dec 30 '16 at 18:00
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In triangle inequality $|z_1+z_2|\leq |z_1|+|z_2|$ we have equality iff $z_1=0$ or there is a $\alpha\geq 0$ with $z_2=\alpha z_1$. Then, $$|z^2-1|=|z|^2+1\Leftrightarrow |z^2+(-1)|=|z^2|+|-1|$$ $$\Leftrightarrow z^2=\alpha (-1)\le 0\Leftrightarrow z=\pm\sqrt{\alpha}i$$ so, $z$ belongs to the imaginary axis.

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