The Question :

Two opposite vertices of a square are represented by complex numbers $9+12i$ and $-5+10i$. Find the complex number representing the other two vertices of the square.

My attempt :

Considering $ABCD$ as the square and $O$ as the origin the known points $A$ and $C$

$OA= 9+12i$

$OC= -5+10i$

By vector addition : $OA=OC+CA$ therefore , $CA=OA-OC$ we get, $CA=9+12i+5-10i=14+2i$

rotating $AC$ using polar for onto $AB$ (Clockwise) $AB=\sqrt{2} CA (\cos 45+ i \sin 45 )$ (45 degrees)





$9+10i+12+16i=OB $

$OB=21+26i$ however the answer in the book is $1+18i$ and $3+4i$


The center of the square is $\frac{A+C}2=2+11i$. So, $MA=7+i$ and you get another vertex $B$ of the square rotating $A$ around $M$ by $\frac\pi2$ radians. That is$$B=M+iMA=1+18i.$$And you'll get the fourts vertex $D$ doing the smae thing, but with an angle of $-\frac\pi2$ radians:$$D=M-iMA=3+4i.$$Your error was to rotate around the origin.

  • $\begingroup$ I see.. Because the square could have been inclined in any direction to the origin , we have to rotate it with respect to a stationary point . $\endgroup$ – Harshit Pandey Aug 29 '17 at 16:02

Here is another way to do this. Let

$$ z_1=9+12i\\ z_3=-5+10i\\ s=\frac{|z3-z2|}{\sqrt{2}},\quad \text{side of the square}\\ \theta=\arg (z3-z2),\quad \text{its angle on the plane} $$


$$ z_2=z_1+se^{i(\theta-\pi/4)}=1+18i\\ z_4=z_1+se^{i(\theta+\pi/4)}=3+4i $$

(I've verified this solution.)


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