# Difficult complex geometry q (1 Viewer)

#### =)(=

##### Active Member
I am pretty lost for ii for some reason visually the modulus looks like R but it cant be

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
I am pretty lost for ii for some reason visually the modulus looks like R but it cant be
Did the sample solution not help?

no not really

#### Lith_30

##### Active Member
I would recommend finding the modulus and the argument separately.

For the modulus you would use pythag
$\bg_white OD^2=OA^2+AD^2$ Given that since OB is a tangent to the circle
$\bg_white \\OD=\sqrt{(OC^2-CA^2)+(R-r)^2}\ \text{pythag again as}\ CO^2=CA^2+OA^2\\OD=\sqrt{(R^2-r^2)+(R-r)^2}\\OD=\sqrt{R^2-r^2+R^2-2Rr+r^2}\\OD=\sqrt{2R^2-2Rr}$ Which is the modulus of D

Now to find the argument, we have to use the triangle COD which is a Isosceles triangle cause CO=CD as they are both the radius of the same circle.

$\bg_white \\\angle{AOC}=90-\theta\ \text{complementary angles}\\\\\therefore\angle{OCA}=\theta\ \text{angle sum of right angle triangle}\\\\\therefore\angle{COD}=\frac{180-\theta}{2}\ \text{Iscosceles triangle}\\\\\angle{COD}=90-\frac{\theta}{2}$
Therefore the angle between line OD and the positive real axis would be $\bg_white \frac{\theta}{2}$ cause it is complementary to $\bg_white \angle{COD}$

Hence the complex number represented by D would be $\bg_white \sqrt{2R^2-2Rr}\times{e^{\frac{i\theta}{2}}}$