If $z_1=iz_2$ and $(z_1-z_3)=i(z_3-z_2)$, prove that $\vert z_3\vert=\sqrt 2 \vert z_1\vert$

Question

If $ z_1=iz_2$ and $(z_1-z_3)=i(z_3-z_2)$, prove that $ |z_3|=\sqrt{2} |z_1|$

Rearranging both given equations and taking Euler form of complex numbers, $$\arg\frac{z_1}{z_2}=\frac{\pi}{2}$$and $$\arg\frac{z_1-z_3}{z_2-z_3}=\frac{3\pi}{2}$$

According to coni's method, if $A(a),B(b),C(c)$ are complex numbers and angle B is $\theta$, then $$\frac{a-b}{c-b}=\frac{|a-b|}{|c-b|}e^{i\theta}$$

Proof: $$a-b=|a-b|e^{i\theta_1}$$ $$c-b=|c-b|e^{i\theta_3}$$ $$\frac{a-b}{c-b}=\frac{|a-b|}{|c-b|}e^{i(\theta_1-\theta_2)}$$ Where $\theta_1-\theta_2$ is the angle between the two vectors.

Applying coni's method, the equations suggest that

1. $z_1$ and $z_2$ are perpendicular,
2. the lengths of $z_1$ and $z_2$ are equal

3. The vectors $z_1-z_3$ and $z_2-z_3$ are perpendicular

   and hence the points $(0,0),z_1,z_3,z_2$ are the vertices of a square where $z_3$ is the diagonal.

Is my approach correct?

Answer 1

Simply put, yes the approach is correct and so is the geometrical interpretation.

For an alternative approach:

You can directly substitute $z_1 = i z_2$ to get:

$$ 2i z_2 = z_3(1+i)$$

Taking modulus on both sides, we get:

$$|z_3| = \sqrt{2} |z_2| = \sqrt{2} |z_1|$$

Answer 2

Substitute $z_1 = iz_2$ into the equation $(z_1-z_3)=i(z_3-z_2)$ and rearrange (solve by $z_3$). Then take the absolute value of $z_3$ by using $|\frac{a}{b}|=\frac{|a|}{|b|}$.