Q:

How to add real number and complex number in polar form

I need to calculate
$$\left| \frac{3}{\sqrt{20}} + i\!\cdot\!\frac{1}{\sqrt{20}}\!\cdot\!e^{i\!\cdot\!\frac{\pi}{3}} \right|$$
Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.
|| is absolute value sign by the way.

A:

It's asking you not for the complex number but its absolute value.
The complement of $x + z$ where $x$ is real is $x + \overline z$.
And the complement of $re^{i\theta}$ is $re^{-i\theta}$.
And $i*e^{i\theta} = e^{i\frac \pi 2}e^{i\theta} = e^{i (\theta +\frac \pi 2)}$.
So $|\frac 3{\sqrt {20}} + i\frac 1{\sqrt{20}}e^{i\frac \pi 3}| =$
$\sqrt{(\frac 3{\sqrt {20}} +\frac 1{\sqrt{20}}e^{i(\frac \pi 3+\frac \pi 2)})(\frac 3{\sqrt {20}} +\frac 1{\sqrt{20}}e^{-i(\frac \pi 3+\frac \pi 2)})}=$
$\sqrt{\frac 9{20} + \frac 3{20}(e^{i\frac {5\pi}6} + e^{-i\frac {5\pi}6}) + \frac 1{20}}=$
$\sqrt{\frac {10}{20} + \frac 3{20}*2\cos \frac {5\pi}6}=$
$\sqrt{\frac 12 - \frac {3\sqrt 3}{20}}$
And I probably made an arithmetic error somewhere....

