\(\displaystyle \bold{(a)}\)
\(\displaystyle 1 + j\sqrt{3} = 2e^{j\frac{\pi}{3}}\)
But
how?
The magic lies in Euler's formula.
\(\displaystyle Ae^{j\theta} = A(\cos \theta + j\sin \theta)\)
The first thing I did to solve this problem was to study the square root, \(\displaystyle \sqrt{3}\).
From basic trigonometry, I know that:
\(\displaystyle \sin 60^{\circ} = \frac{\sqrt{3}}{2}\)
Then, I did the same angle for the cosine too:
\(\displaystyle \cos 60^{\circ} = \frac{\sqrt{1}}{2} = \frac{1}{2}\)
This gave me an idea that the angle could be \(\displaystyle 60^{\circ} = \frac{\pi}{3}\).
\(\displaystyle Ae^{j\frac{\pi}{3}} = A\left(\cos \frac{\pi}{3} + j\sin \frac{\pi}{3}\right) = A\left(\frac{1}{2} + j\frac{\sqrt{3}}{2}\right)\)
After that it was so obvious that \(\displaystyle A\) should be \(\displaystyle 2\) to get the required form.
\(\displaystyle 2e^{j\frac{\pi}{3}} = 2\left(\frac{1}{2} + j\frac{\sqrt{3}}{2}\right) = 1 + j\sqrt{3}\)
So, basically I did the calculations in reverse!
Another way to solve the problem
without thinking too much is to do this:
\(\displaystyle A = \sqrt{1^2 + \left(\sqrt{3}\right)^2} = \sqrt{1 + 3} = \sqrt{4} = 2\)
\(\displaystyle \theta = \tan^{-1}\frac{\sqrt{3}}{1} = \frac{\pi}{3}\)
