On the Argand plane $z_1, z_2$ and $z_3$ are respectively the vertices of an isosceles triangle $\triangle ABC$ with $AC = BC$ and equal angles are $θ$. If $z_4$ is the incentre of the triangle then prove that $(z_2 – z_1 ) (z_3 – z_1 ) = (1 + \sec θ) (z_4 – z_1 )^2$
I applied Coni's method of rotation of unit vectors (involving complex numbers).
$$(z_2-z_1)/|z_2-z_1|e^{i\theta/2}=(z_4-z_2)/|z_4-z_2|$$ and $$(z_1-z_2)/|z_1-z_2|e^{i\theta/2}=(z_1-z_4)/|z_1-z_4|$$
After that what to do?
We have $$\arg\left(\frac{z_2-z_1}{z_4-z_1}\right)=-\arg\left(\frac{z_3-z_1}{z_4-z_1}\right)\tag1$$
Also, applying the law of sines to $\triangle{ACI}$ where $I(z_4)$ gives $$\frac{|z_4-z_1|}{\sin\left(\frac{\pi}{2}-\theta\right)}=\frac{|z_3-z_1|}{\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)}\tag2$$
From $(1)(2)$, $$\begin{align}\frac{z_2-z_1}{z_4-z_1}\cdot\frac{z_3-z_1}{z_4-z_1}&=\left|\frac{z_2-z_1}{z_4-z_1}\right|\left|\frac{z_3-z_1}{z_4-z_1}\right|\tag3\\\\&=2\cos\frac{\theta}{2}\cdot\frac{\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)}{\sin\left(\frac{\pi}{2}-\theta\right)}\\\\&=\frac{2\cos^2\frac{\theta}{2}}{\cos\theta}\\\\&=\frac{\cos\theta+1}{\cos\theta}\\\\&=1+\sec\theta\end{align}$$
Added :
Using your idea, we get $$\frac{z_2-z_1}{|z_2-z_1|}e^{\pm\frac{i\theta}{2}}=\frac{z_4-z_1}{|z_4-z_1|}=\frac{z_3-z_1}{|z_3-z_1|}e^{\mp\frac{i\theta}{2}}$$ from which we have $(3)$. The result follows from this.