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There are two complex numbers $z_1$ and $z_2$ related by $\frac{z_1}{z_2} = i$, which implies that they have an angle of $\frac{\pi}{2}$ between them ($e^{i\frac{\pi}{2}} = i$).

Does the way in which we measure the angle matter? To clarify what I mean, let me assume $z_1 = 1+i$ and $z_2 = 1-i$. This gives us two complex numbers in the first and fourth quadrants. The angle between these two could be either $\frac{\pi}{2}$ or $\frac{3\pi}{2}$. For this pair of numbers, $\frac{z_1}{z_2}$ turns out to be $i$.

If we now assume $z_1 = -1+i$ and $z_2 = -1-i$, two complex numbers in the second and third quadrants. The angle between these two could again be either $\frac{\pi}{2}$ or $\frac{3\pi}{2}$. But $\frac{z_1}{z_2}$ in this case turns out to be $-i$.

What is the difference in these cases? Is it because we measure arguments from the real axis?

This was the original question: If $z_1=iz_2$ and $(z_1-z_3)=i(z_3-z_2)$, prove that $\vert z_3\vert=\sqrt 2 \vert z_1\vert$ and I realized that $z_3$ could be either the fourth vertex of a square with $(0,0), z_1, z_2$ or $z_3$ could be $(0,0)$ itself depending upon the sign of the relation: $\frac{z_1-z_3}{z_3-z_2} = i$ or $-i$ respectively.

zxen
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1 Answers1

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It's because angles are measured counterclockwise from the positive real axis. That is, in the polar representation of a complex number $z = r e^{i\theta}$, $\theta$ is (up to a multiple of $2\pi$) the angle counterclockwise from the positive real axis to the ray from the origin through $z$. And if $z = r e^{i\theta}$ and $w = r' e^{i\theta'}$, $z/w = (r/r') e^{i (\theta - \theta')}$ where $\theta - \theta'$ is the angle counterclockwise from the ray through $w$ to the ray through $z$.

Robert Israel
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