How to add real number and complex number in polar form

Question

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.

Answer 1

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....

Answer 2

In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $\pi$.

When the arguments are the same, $\ r_1e^{i\theta} \pm r_2{i\theta} = (r_1 \pm r_2)e^{i\theta}$

When the arguments are separated by $\pi$, $\ r_1e^{i\theta} \pm r_2e^{i(\theta \pm \pi)} = r_1e^{i\theta} \mp r_2e^{i\theta} = (r_1 \mp r_2)e^{i\theta}$