# How do you multiply  (1+6i)(2+2i)  in trigonometric form?

Dec 30, 2015

First of all we have to convert these two numbers into trigonometric forms.
If $\left(a + i b\right)$ is a complex number, $u$ is its magnitude and $\alpha$ is its angle then $\left(a + i b\right)$ in trigonometric form is written as $u \left(\cos \alpha + i \sin \alpha\right)$.
Magnitude of a complex number $\left(a + i b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(1 + 6 i\right)$ and $\theta$ be its angle.
Magnitude of $\left(1 + 6 i\right) = \sqrt{{1}^{2} + {6}^{2}} = \sqrt{1 + 36} = \sqrt{37} = r$
Angle of $\left(1 + 6 i\right) = T a {n}^{-} 1 \left(\frac{6}{1}\right) = {\tan}^{-} 1 \left(6\right) = \theta$

$\implies \left(1 + 6 i\right) = r \left(C o s \theta + i \sin \theta\right)$

Let $s$ be the magnitude of $\left(2 + 2 i\right)$ and $\phi$ be its angle.
Magnitude of $\left(2 + 2 i\right) = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{4 + 4} = \sqrt{8} = s$
Angle of $\left(2 + 2 i\right) = T a {n}^{-} 1 \left(\frac{2}{2}\right) = T a {n}^{-} 1 \left(1\right) = \phi$

$\implies \left(2 + 2 i\right) = s \left(C o s \phi + i \sin \phi\right)$

Now,
$\left(1 + 6 i\right) \left(2 + 2 i\right)$
$= r \left(C o s \theta + i \sin \theta\right) \cdot s \left(C o s \phi + i \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi + i \sin \theta \cos \phi + i \cos \theta \sin \phi + {i}^{2} \sin \theta \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi - \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi + \cos \theta \sin \phi\right)$
$= r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$
Here we have every thing present but if here directly substitute the values the word would be messy for find $\theta + \phi$ so let's first find out $\theta + \phi$.

$\theta + \phi = {\tan}^{-} 1 \left(6\right) + {\tan}^{-} 1 \left(1\right)$
We know that:
${\tan}^{-} 1 \left(a\right) + {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a + b}{1 - a b}\right)$
$\implies {\tan}^{-} 1 \left(6\right) + {\tan}^{-} 1 \left(1\right) = {\tan}^{-} 1 \left(\frac{6 + 1}{1 - 6 \cdot 1}\right)$
$= {\tan}^{-} 1 \left(\frac{7}{- 5}\right) = {\tan}^{-} 1 \left(- \frac{7}{5}\right)$
$\implies \theta + \phi = {\tan}^{-} 1 \left(- \frac{7}{5}\right)$

$r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$
$= \sqrt{37} \sqrt{8} \left(\cos \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right)\right)$
$= \sqrt{296} \left(\cos \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right)\right)$

$\left(1 + 6 i\right) \left(2 + 2 i\right) = 2 + 2 i + 12 i + 12 {i}^{2} = 2 + 14 i - 12 = - 10 + 14 i$
Now change $- 10 + 14 i$ in trigonometric form.
Magnitude of $- 10 + 14 i = \sqrt{{\left(- 10\right)}^{2} + {\left(14\right)}^{2}} = \sqrt{100 + 196} = \sqrt{296}$
Angle of $- 10 + 14 i = {\tan}^{-} 1 \left(\frac{14}{-} 10\right) = {\tan}^{-} 1 \left(- \frac{7}{5}\right)$
$\implies - 10 + 14 i = \sqrt{296} \left(\cos \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{7}{5}\right)\right)\right)$