# How do you write the following in trigonometric form and perform the operation given (sqrt3+i)(1+i)?

Apr 30, 2018

$\left(\setminus \sqrt{3} + i\right) \left(1 + i\right) = 2 \setminus \textrm{c i s} \left(\setminus \frac{\pi}{6}\right) \setminus \sqrt{2} \setminus \textrm{c i s} \left(\frac{\pi}{4}\right)$

$= 2 \setminus \sqrt{2} \setminus \textrm{c i s} \left(\frac{\pi}{6} + \frac{\pi}{4}\right) = 2 \setminus \sqrt{2} \setminus \textrm{c i s} \left(\frac{5 \pi}{12}\right)$

#### Explanation:

I don't really get these questions. These are easy numbers to multiply in rectangular form -- why make it harder ?

The two factors are the two biggest cliches in trig, 30/60/90 and 45/45/90, so it's OK to just write down the trigonometric forms and multiply without too much thought, as I did above.

Let's turn each factor to trigonometric form, which is

$z = r \left(\cos \theta + i \sin \theta\right) = r \setminus \textrm{c i s} \setminus \theta ,$

basically polar coordinates, but still a rectangular form.

Let's remember how to turn a number into trigonometric or polar form:

$a + b i = r \left(\cos \theta + i \sin \theta\right)$

We want to solve $a = r \cos \theta$, $b = r \sin \theta$. We get

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\theta = \textrm{\arctan 2} \left(b / , a\right)$

Usually folks write a plain arctan for the angle, but I want to make clear we need to two parameter, four quadrant inverse tangent to properly convert to polar. The odd $/ ,$ notation reminds us it's two parameters and which is which. It's not much of an issue in this question as both factors are in the first quadrant, positive sine and cosine.

$| \sqrt{3} + i | = \setminus \sqrt{3 + 1} = 2 , \angle \left(\sqrt{3} + i\right) = \textrm{\arctan 2} \left(1 / , 3\right) = \frac{\pi}{6}$

$| 1 + i | = \setminus \sqrt{1 + 1} = \sqrt{2} , \angle \left(1 + i\right) = \textrm{\arctan 2} \left(1 / , 1\right) = \frac{\pi}{4}$

$\left(\setminus \sqrt{3} + i\right) \left(1 + i\right) = \left(2 \setminus \textrm{c i s} \left(\frac{\pi}{6}\right)\right) \left(\sqrt{2} \setminus \textrm{c i s} \left(\setminus \frac{\pi}{4}\right)\right)$

$= 2 \sqrt{2} \setminus \textrm{c i s} \left(\frac{\pi}{6} + \setminus \frac{\pi}{4}\right)$

$= 2 \sqrt{2} \setminus \textrm{c i s} \left(\frac{5 \pi}{12}\right)$

We'll leave it there, but we know that equals

$\left(\setminus \sqrt{3} + i\right) \left(1 + i\right) = \left(\setminus \sqrt{3} - 1\right) + i \left(\setminus \sqrt{3} + 1\right)$