# How do you multiply  (2-3i)(-3-7i)  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 $\left(2 - 3 i\right)$ and $\theta$ be its angle.
Magnitude of $\left(2 - 3 i\right) = \sqrt{{2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = r$
Angle of $\left(2 - 3 i\right) = T a {n}^{-} 1 \left(- \frac{3}{2}\right) = \theta$

$\implies \left(2 - 3 i\right) = r \left(C o s \theta + i \sin \theta\right)$

Let $s$ be the magnitude of $\left(- 3 - 7 i\right)$ and $\phi$ be its angle.
Magnitude of $\left(- 3 - 7 i\right) = \sqrt{{\left(- 3\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{9 + 49} = \sqrt{58} = s$
Angle of $\left(- 3 - 7 i\right) = T a {n}^{-} 1 \left(\frac{- 7}{-} 3\right) = T a {n}^{-} 1 \left(\frac{7}{3}\right) = \phi$

$\implies \left(- 3 - 7 i\right) = s \left(C o s \phi + i \sin \phi\right)$

Now,
$\left(2 - 3 i\right) \left(- 3 - 7 i\right)$
$= r \left(C o s \theta + i \sin \theta\right) \cdot s \left(C o s \phi + i \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi + i \sin \theta \cos \phi + i \cos \theta \sin \phi + {i}^{2} \sin \theta \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi - \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi + \cos \theta \sin \phi\right)$
$= r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$
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 $\theta + \phi$.

$\theta + \phi = {\tan}^{-} 1 \left(- \frac{3}{2}\right) + {\tan}^{-} 1 \left(\frac{7}{3}\right)$
We know that:
${\tan}^{-} 1 \left(a\right) + {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a + b}{1 - a b}\right)$
$\implies {\tan}^{-} 1 \left(- \frac{3}{2}\right) + {\tan}^{-} 1 \left(\frac{7}{3}\right) = {\tan}^{-} 1 \left(\frac{\left(- \frac{3}{2}\right) + \left(\frac{7}{3}\right)}{1 - \left(- \frac{3}{2}\right) \left(\frac{7}{3}\right)}\right)$
$= {\tan}^{-} 1 \left(\frac{- 9 + 14}{6 + 21}\right) = {\tan}^{-} 1 \left(\frac{5}{27}\right)$
$\implies \theta + \phi = {\tan}^{-} 1 \left(\frac{5}{27}\right)$

$r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$
$= \sqrt{13} \sqrt{58} \left(\cos \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right)\right)$
$= \sqrt{754} \left(\cos \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right)\right)$

$\left(2 - 3 i\right) \left(- 3 - 7 i\right) = - 6 - 14 i + 9 i + 21 {i}^{2} = - 6 - 5 i - 21 = - 27 - 5 i$
Now change $- 27 - 5 i$ in trigonometric form.
Magnitude of $- 27 - 5 i = \sqrt{{\left(- 27\right)}^{2} + {\left(- 5\right)}^{2}} = \sqrt{729 + 25} = \sqrt{754}$
Angle of $- 27 - 5 i = {\tan}^{-} 1 \left(- \frac{5}{-} 27\right) = {\tan}^{-} 1 \left(\frac{5}{27}\right)$
$\implies - 27 - 5 i = \sqrt{754} \left(\cos \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{5}{27}\right)\right)\right)$