If $z_1+z_2+z_3=z_1z_2z_3$, then $z_1$, $z_2$ and $z_3$ cannot lie all above the real axis.

complex numbers

I am trying to solve the following problem:

If $z_1$, $z_2$ and $z_3$ are complex numbers such that $z_1+z_2+z_3=z_1z_2z_3$, then $z_1$, $z_2$ and $z_3$ cannot lie all above the real axis.

It is a problem in Limits by Alan Beardon that I have been self-studying. I have been stuck on this problem for quite some time.

Writing $z_j=x_j+iy_j$ with $x_j,y_j$ real numbers for $j\in\{1,2,3\}$, we have

$$
z_1+z_2+z_3 = (x_1+x_2+x_3)+i(y_1+y_2+y_2)
$$

and

$$
z_1z_2z_3 = (x_1x_2x_3-x_1y_2y_3-y_1x_2y_3-y_1y_2x_3)
+i(x_1x_2y_3+x_1y_2x_3+y_1x_2x_3-y_1y_2y_3)
$$

so $x_1+x_2+x_3=x_1x_2x_3-x_1y_2y_3-y_1x_2y_3-y_1y_2x_3$ and $y_1+y_2+y_3=x_1x_2y_3+x_1y_2x_3+y_1x_2x_3-y_1y_2y_3$.

But I am not sure how it can help to show that at least one of the $y_j's$ is not positive. Most of the other problems in this section are about various forms of triangular inequality, but I don't know how to use it here.

Any idea?

Best Answer

Look at the arguments (I constrain them to the interval $[0,2\pi)$),

If $z_1z_2z_3=z_1+z_2+z_3$, and they all have positive imaginary parts, then

  • the arguments of all three are in the interval $(0,\pi)$,
  • the argument of their average is also in that interval, and exceeds the smallest argument of the three numbers, and is also smaller than the largest of the three arguments,
  • but the argument of the product is either larger than any of the three, and (in case it wraps around $2\pi$) less than any of three $+2\pi$.
Related Question