Not able to understand clockwise/anti-clockwise nature of the angles of triangle

Question

If $A(z_1)$, $B(z_2)$, $C(z_3)$ are the vertices of an equilateral triangle ABC, find the value of $\arg\left(\frac{z_2+z_3-2z_1}{z_3-z_2}\right)$

The numerator can be written as $(z_2-z_1)+(z_3-z_1)$

Now, we can write both these terms in terms of the denominator i.e. $z_3-z_2$ using rotation theorem.

But I wonder whether to use $e^{i\frac\pi3}$ or $e^{-i\frac\pi3}$

Can you help? Thanks.

Answer 1

You should use both. What I mean is there are two cases.

• $z_1-z_2=(z_3-z_2)e^{i\frac\pi3}$, i.e., $z_1-z_2$ is $z_3-z_2$ rotated counterclockwise $\frac\pi3$.
  That means $z_1=\frac{1-i\sqrt3}2z_2+\frac{1-i\sqrt3}2z_3$. Replacing $z_1$, we get $$\frac{z_2+z_3-2z_1}{z_3-z_2}=\frac{i\sqrt3z_2-i\sqrt3z_3}{z_3-z_2}=-i\sqrt3$$ The wanted value is $\frac{-\pi}2$.
• $z_1-z_2=(z_3-z_2)e^{-i\frac\pi3}$, i.e., $z_1-z_2$ is $z_3-z_2$ rotated clockwise $\frac\pi3$.
  That means $z_1=\frac{1+i\sqrt3}2z_2+\frac{1-i\sqrt3}2z_3$. Replacing $z_1$, we get $$\frac{z_2+z_3-2z_1}{z_3-z_2}=\frac{-i\sqrt3z_2+i\sqrt3z_3}{z_3-z_2}=i\sqrt3$$ The wanted value is $\frac{\pi}2$.

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More generally, if $A(z_1)$, $B(z_2)$, $C(z_3)$ are the vertices of an isosceles triangle $ABC$ with $AB=AC$, then $\arg\left(\frac{z_2+z_3-2z_1}{z_3-z_2}\right)=\pm\frac\pi2$.