I need to show that if $\arg\left(\frac{z_1+z_2}{z_1-z_2}\right)=\frac{\pi}{2}$ then $|z_1|=|z_2|$. How should I work this out? I know that $\arg\left(\frac{z_1+z_2}{z_1-z_2}\right) = \arg(z_1+z_2)-\arg(z_1-z_2)$ and since the argument is $\frac{\pi}{2}$ than the complex number lies on the positive imaginary axis. Is this correct?
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$T(z) = (1+z)/(1-z)$ is a Möbius transformation with $T(-1) = 0$, $T(1) = \infty$ and $T(i) = i$ and therefore mapping the unit circle onto the (extended) imaginary axis.
So $ \arg T(z) = \pi/2 $ implies $ |z| = 1 $. Now set $ z = z_2/z_1$.
(A possible geometric argument: $ \arg(1+z) - \arg(1-z) = \pi/2$ implies that the triangle with the points $(1, -1, z)$ has a right angle at $z$. Using the Converse of Thales Theorem if follows that $z$ lies on the circle whose diameter is the hypothenuse $[-1, 1]$.)
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$\begingroup$ @user120768: $\arg T(z) = \pi/2$ means that $T(z)$ is on the imaginary axis. Therefore $z$ is on the unit circle. A Möbius transform is bijective and maps circles or lines on circles or lines. $\endgroup$ – Martin R Dec 10 '14 at 23:03
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$\begingroup$ but for example z=3i is also on the imaginary axis but |z| is not 1 $\endgroup$ – user120768 Dec 10 '14 at 23:06
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$\begingroup$ @user120768: If T (z) lies on the imaginary axis then |z| = 1. T(3i) does not lie on the imaginary axis. $\endgroup$ – Martin R Dec 10 '14 at 23:07
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Yes, this is correct.
Let $z_1=u+iv$, $z_2=r+is$ and equal the real part of the quotient to $0$: you'll be done in no time.
