Explanation that sin(x+iy) is one-to-one

[hotvette]

TL;DR Summary

Complex Variables by Stephen Fisher

Not hw, just reading the textbook. In section 1.5, page 50, the book goes through an explanation that $\sin(x+iy)$ is one-to-one if $0 \le x < \pi/2$ and $y \ge 0$. At one point the book states that for $1 = -e^{-i x_1}\,e^{-i x_2}\,e^{y_1}\,e^{y_2}$ the absolute value of the left side is 1 and that of the right side is $e^{y_1 + y_2}$. It then states that this result implies that $-1 = e^{-ix_1-ix_2}$.

I don't at all see why $\left\vert-e^{-i x_1}\,e^{-i x_2}\,e^{y_1}\,e^{y_2}\right\vert = e^{y_1 + y_2}$. Can someone explain?

 

[Infrared]

Just simplify $|-e^{-ix_1}e^{-ix_2}e^{y_1}e^{y_2}|=|-e^{-ix_1}|\cdot |e^{-ix_2}|\cdot |e^{y_1}|\cdot |e^{y_2}|=1\cdot 1\cdot e^{y_1}\cdot e^{y_2}=e^{y_1+y_2}.$

Keep in mind that if $t$ is real, then $|e^{it}|=|\cos(t)+i\sin(t)|=1$.

 

[hotvette]

I figured it must be simple. I too often miss the obvious. Thanks!