Question: Show that $(1/z_1)+(1/z_2)+(1/z_3)=1$ given that $z_1$, $z_2$, $z_3$ are complex satisfying $z_1z_2z_3=1$, $z_1+z_2+z_3=1$ and $|z_1|=|z_2|=|z_3|=1$. Also compute $z_1$, $z_2$ and $z_3$.

So firstly I started out by trying to compute $z_1$ $z_2$ and $z_3$ by splitting them into real and imaginary parts - e.g. $z_1=x_1+iy_1$ - and then using the facts given in the question to get a set of simultaneous equations for $x_1,x_2,x_3,y_1,y_2,y_3$. I immediately realised this was the wrong approach as i ended up with some hideous looking equations that were unsolvable (or beyond my expertise at least!).

I then used the first two equations to show that: $$ 1=\frac{1}{z_2z_3}+\frac{1}{z_1z_3}+\frac{1}{z_1z_2} $$

I know that I must now use the final equation (with the moduli) to show that $$ z_1=z_2z_3,\quad z_2=z_1z_3,\quad z_3=z_1z_2 $$ but I am dreadfully stuck on this part so any sort of help or advice would be much appreciated.

P.S. I have also tried to us the polar form of a complex number but ran into similar difficulties as the first approach.

  • $\begingroup$ $Z_1Z_2=Z_3$ and the other equations are not necessarily true. Take $(Z_1,Z_2,Z_3)=(1,i,-i)$, for example. $\endgroup$ – Arnaud D. Oct 7 '15 at 16:32
  • $\begingroup$ Use $1/z = \bar z$ if $|z|=1$. $\endgroup$ – A.S. Oct 7 '15 at 16:40
  • $\begingroup$ I can't believe I'd been so silly as to overlook using the conjugate. Thank you for your insightful help this was much appreciated $\endgroup$ – WillBee Oct 7 '15 at 17:12

Given $|z_{1}| = |z_{2}| = |z_{3}| = 1\Rightarrow z_{1}\cdot \bar{z_{1}} = 1$ and $$\displaystyle z_{2}\cdot \bar{z_{2}} = 1$$ and $$\displaystyle z_{3}\cdot \bar{z_{3}} = 1$$

Also given $$\displaystyle z_{1}+z_{2}+z_{3} = 1...............(1)$$ and $$\displaystyle z_{1}\cdot z_{2}\cdot z_{3} = 1$$

Now taking Conjugate on both side of first equation

We get $$\displaystyle \bar{z_{1}}+\bar{z_{2}}+\bar{z_{3}} = 1\Rightarrow \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = 1$$ (From first line.)

So we get $$z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1} = 1$$

Now we will form an cubic equation in terms of $t\;,$ Whose roots are $t=z_{1}\;, t=z_{2}\;,t=z_{3}$

So we get $$\displaystyle (t-z_{1})(t-z_{2})(t-z_{3})=0\Rightarrow t^3-(z_{1}+z_{2}+z_{3})t^2+(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})t-z_{1}\cdot z_{2}\cdot z_{3}=0$$

So we get $$t^3-t^2+t-1 = 0\Rightarrow t^2(t-1)+1(t-1) = 0\Rightarrow (t^2+1)(t-1) =0$$

So we get $$t=i\;-i,\;1$$

So we get $z_{1},z_{2},z_{3}$ is all possible triplets of $i,-i,1$


If $|Z_1|=1$, then $\overline{Z_1}=Z_1^{-1}$, and similarly for $Z_2,Z_3$. So \begin{align*}\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3} & =\overline{Z_1}+\overline{Z_2}+\overline{Z_3} \\ & = \overline{Z_1+Z_2+Z_3} \\ & = 1.\end{align*}

Now define a polynomial $P\in \mathbb{C}[X]$ by $P=(X-Z_1)(X-Z_2)(X-Z_3)$. Developping the product and using all the given equations yields $P=X^3-X^2+X-1$. But$$X^3-X^2+X-1=X^2(X-1)+(X-1)=(X^2+1)(X-1)=(X-1)(X-i)(X+i).$$

So by comparing the two decompositions of $P$ one gets that the only solution is $\{Z_1,Z_2,Z_3\}=\{1, i, -i\}$.


WLOG let $Z_r=\cos A_r+i\sin A_r, 1\le r\le3$

$1=Z_1+Z_2+Z_3=\sum\cos A_r+i(\sum\sin A_r)$

$0=\sin A_1+\sin A_2+\sin A_3$

$1=\cos A_1+\cos A_2+\cos A_3 $

$\dfrac1{Z_1}=\cos A_r-i\sin A_r$

$\sum\dfrac1{Z_r}=\sum\cos A_r-i\sum\sin A_r=?$


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