# Proving complex number equality given that $a\in \mathbb{R}$

Given that $$a \in \mathbb{R}$$ and $$z \in \mathbb{C}^*$$ such that $$a=|z+\frac{1}{z}|$$ and $$|z|_m$$ and $$|z|_M$$ denotes the minimum and the maximum value of $$|z|$$ respectively, show that: $$(|z|_m+|z|_M)^2=a^2+4$$.

I came across this problem in a periodical. I tried to solve it, but something is odd for me atleast. $$|z|_m$$ is 0 if $$z=\pm i$$, but if so, then the LHS is just $$|z|_M^2$$, and $$|z|_m$$ and $$|z|_M$$ have to be different, otherwise it would have no use to talk about minimum and maximum whatsoever. If so far my argumentation is correct, then this equality won't be true. Have I overlooked something?

• What do you mean by minimum and maximum values of $|z|$?
– Mark
Commented Apr 6 at 11:14
• @Mark this is my problem too. This is how the problem says it. My best guess is that if we were to graph $|z+\frac{1}{z}|$ what would be its minimum and maximum, but it lead nowhere. Commented Apr 6 at 11:17
• @Mark: I am guessing it means maximum and minimum values of $|z|$ among all $z\in\mathbb C$ satisfying $\left|z+\frac1z\right|=a$. Commented Apr 6 at 11:53
• I read it as $a$ is as fixed value in $\mathbb R$ and that $z$ is a variable value in $\mathbb C$ so that $|z+\frac 1z| = a$ then $|z|_m, |z|_M$ are the min and max possible value of $|z|$. Commented Apr 6 at 15:36
• If $z =\pm i$ then the LHS is not $|z|_M^2$ it is $4|z|_M^2$ and why do you say that the equality would not hold. Why can't we have $4|z|_M^2 =a^2 + 4$. If $z =\pm i$ then $|z|_m = |z|_m = |z| = 1$. And $\frac 1z =\frac 1{\pm i} =\mp i$ and $z + \frac 1z = 0$. So $a = 0$. So the LHS is $4|z|^2 = 4$ and the RHS is $a^2 + 4=0^2 + 4 = 4$. It holds. Why did you think it wouldn't Commented Apr 6 at 15:47

## 2 Answers

If $$a=\bigg| z+ \frac {1}{z} \bigg|$$, we can assume $$z=R(\cos x + i\sin x)$$ and $$z^{-1}=R^{-1}(\cos x - i \sin x)$$, where $$R=|z|$$.

Now $$a=\bigg|R(\cos x+ i\sin x) + \frac{1}{R}(\cos x - i\sin x)\bigg|$$ Or $$a=\bigg|\cos x\bigg(R+\frac{1}{R}\bigg) + i \sin x\bigg(R-\frac{1}{R}\bigg)\bigg|$$ Or $$a^2=\cos^2 x\bigg(R+\frac{1}{R}\bigg)^2+\sin^2x\bigg(R-\frac{1}{R}\bigg)^2$$ Or $$a^2= R^2 + \frac{1}{R^2} + 2\cos^2x-2\sin^2x$$ Or $$a^2= R^2 + \frac{1}{R^2} + 4\cos^2x-2$$ Or $$a^2-4\cos^2 x = \bigg(R-\frac{1}{R}\bigg)^2 \implies a^2≥\bigg(R-\frac{1}{R}\bigg)^2$$

Applying inequality laws, We can say $$-a≤ R - \frac{1}{R} ≤ a$$ At $$x=90°$$, $$R-\frac{1}{R}=a \implies R=\frac{a±\sqrt{a^2 +4}}{2}=R_{max}$$ Similarly at $$x=270°$$, $$-a=R-\frac{1}{R} \implies R=\frac{-a±\sqrt{a^2 +4}}{2}=R_{min}$$

And hence $$(R_{max}+R_{min})^2=a^2+4$$[QED]

• I think I am missing something. $|x|²=|x²|$, thus wouldn't we need the product of the two terms when squaring $|(R+\frac{1}{R})\cos{x}+i(R-\frac{1}{R})\sin{x}|$? Commented Apr 7 at 5:45
• It's not squaring, it's finding out the modulus. $|z|=|(R+\frac{1}{R})\cos{x}+i(R-\frac{1}{R})\sin{x}|$ Modulus of $(x+iy)=\sqrt{x^2+y^2}$ It is actually finding the distance of $(x,y)$ on Argand plane from the origin. So $$|z|^2=((R+\frac{1}{R})\cos{x})^2+((R-\frac{1}{R})\sin{x})^2$$ @fikooo
– Gwen
Commented Apr 7 at 5:52
• It is...my bad...you are right. I was exhausted in the morning. Commented Apr 7 at 10:37
• by the way why isn't $R_min=\frac{-a-sqrt{a^2+4}}{2}$? It seems to me that it is smaller than the other value. Commented Apr 10 at 16:24
• @fikooo I forgot to put $±$. It doesn't matter though
– Gwen
Commented Apr 10 at 18:13

If $$z=\pm i$$ are solutions to this equation then it must be that $$a=0$$, and then the equation $$\left|z+\frac{1}{z}\right|=0 \tag{1}\label{eq:1}$$ has only $$\pm i$$ as solutions. Therefore the maximum and minimum absolute values of solutions to equation \eqref{eq:1} are 1, so, by definition, $$|z|_m=1=|z|_M$$, and the required equation holds since $$(1+1)^2=0+4$$.