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Show that the real part of any solution of $(z+1)^{100}=(z-1)^{100}$ must be zero.

I know that I must use the fact $e^{2\pi ki}$ somewhere but not sure exactly how or where to use it.

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3 Answers 3

up vote 1 down vote accepted

If $z=x+iy$ and if we put the value of $z$ in $|z+1|=|z-1|$ which has been derived in other answers,

$$|x+iy+1|=|x+iy-1|\implies (x+1)^2+y^2=(x-1)^2+y^2$$ $$\implies (x+1)^2-(x-1)^2=0\implies4\cdot x\cdot1=0\implies x=0$$

A little generalization:

Let $|z-w|=|z+w|$

If $z=x+iy,w=a+ib$ where $x,y,a,b$ are real and $a\cdot b\ne0$

$$(x-a)^2+(y-b)^2=(x+a)^2+(y+b)^2\implies x\cdot a+y\cdot b=0$$

If $b=0\implies w=a,x\cdot a=0\implies x=0$ as $a\ne0\implies z=iy$ (In our case, $a=1$)

If $a=0\implies w=ib,y\cdot b=0\implies y=0$ as $b\ne0\implies z=x$

Geometrically, $$\arg(z)-\arg(w)=\arctan\frac yx-\arctan\frac ba$$ $$=\arctan \left(\frac{\frac yx-\frac ba}{1+\frac yx\cdot\frac ba}\right)=\arctan\left(\frac{ay-bx}{x\cdot a+y\cdot b}\right)=\frac\pi2 $$

$$\implies z\perp w $$

If $b=0,\arg(w)=\arctan 0=0$ or $\pi,$ so $w$ is parallel to $X$ axis, $z$ being perpendicular to $w,$ will be parallel to $Y$ axis. Observe that, $\arg(z)=\frac\pi2+0=\frac\pi2$ or $\frac\pi2+\pi\equiv-\frac\pi2$

Similarly for $a=0$

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Thanks a lot, Lab! –  Q.matin Jan 27 '13 at 6:05
@Q.matin, my pleasure. Please find the edited version. –  lab bhattacharjee Jan 27 '13 at 13:05
Even better thanks!! Going to make this the accepted answer, thanks again! –  Q.matin Jan 27 '13 at 22:43

Take the modulus of each side: $|z+1|^{100} = |z-1|^{100}$.

Hence $|z+1| = |z-1|$ which means (interpret the modulus geometrically) that $z$ is on the perpendicular bisector of $-1$ and $1$.

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Thanks a lot, I do understand interpreting it geometrically which will give that are perpendicular bisector but I do not understand how the powers vanished when you took the modulus? –  Q.matin Jan 27 '13 at 1:14
They didn't vanish but $x\mapsto x^{100}$ is a bijection from $\mathbb{R}^+$ to $\mathbb{R}^+$ so $a^{100}=b^{100}$ for $a\ge 0, b\ge 0$ implies $a=b$ –  xavierm02 Jan 27 '13 at 1:27
Thanks a lot!!! –  Q.matin Jan 27 '13 at 2:08
+1 Very nice and elementary –  DonAntonio Jan 27 '13 at 2:43

Notice that the equation gives $|z+1|^{100} = |z-1|^{100}$, which gives $|z-1|=|z+1|$. Squaring gives $|z+1|^2 = |z|^2+z +\overline{z} +1 = |z|^2-z -\overline{z} +1 = |z-1|^2$. This gives $z +\overline{z}=0$, from which it follows that $\text{Re } z = 0$.

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Thanks a lot! But how did the powers of 100 vanish when you took the modulus? –  Q.matin Jan 27 '13 at 1:15
They didn't vanish, I took the $100$th root of both sides. The function $f: [0,\infty) \to [0,\infty)$ given by $f(x) = x^{100}$ is a bijection. Hence if $x,y \geq 0$, then $x=y$ iff $f(x) = f(y)$ iff $x^{100} = y^{100}$. –  copper.hat Jan 27 '13 at 1:52
Haven't been taught bijection yet. Seems very important for this types of problems. Going to look it up. Thanks a lot! –  Q.matin Jan 27 '13 at 2:08
The important part of being a bijection here is that if $f(x) = f(y)$ (ie, $|z+1|^{100} = |z-1|^{100}$), then $x=y$ (ie, $|z+1|=|z-1|$). –  copper.hat Jan 27 '13 at 4:14

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