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?

  • 1
    $\begingroup$ Are you aware that you can format your maths with MathJaX in the title of your question, just like you would/did in the body of your question. $\endgroup$ – gebruiker Dec 12 '15 at 13:16
  • $\begingroup$ Someone said that there will be problems while searching for a question if I used latex on title. $\endgroup$ – Aditya Dev Dec 12 '15 at 13:19
  • $\begingroup$ I have never heard of "coni's method of rotation", is that a standard term? $\endgroup$ – Martin R Dec 12 '15 at 13:20
  • $\begingroup$ That's what my teacher told me. $\endgroup$ – Aditya Dev Dec 12 '15 at 13:21
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    $\begingroup$ Latex only titles are difficult to search for, yes. That is why you should alway use some text in your tilte as well. However the parts of your title that is maths, should be formatted like everyting else on this site. Use only single $'s though (inline style). The $$-environment (which renders displaystyle) should not be used in titles. $\endgroup$ – gebruiker Dec 12 '15 at 13:25

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:

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


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 $|\frac{a}{b}|=\frac{|a|}{|b|}$.

  • 3
    $\begingroup$ Is approach correct? $\endgroup$ – Aditya Dev Dec 12 '15 at 13:19

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