# How do you multiply  (4+6i)(3+7i)  in trigonometric form?

##### 1 Answer
Jan 10, 2016

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 $\left(4 + 6 i\right)$ and $\theta$ be its angle.
Magnitude of $\left(4 + 6 i\right) = \sqrt{{4}^{2} + {6}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13} = r$
Angle of $\left(4 + 6 i\right) = T a {n}^{-} 1 \left(\frac{6}{4}\right) = {\tan}^{-} 1 \left(\frac{3}{2}\right) = \theta$

$\implies \left(4 + 6 i\right) = r \left(C o s \theta + i \sin \theta\right)$

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

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

Now,
$\left(4 + 6 i\right) \left(3 + 7 i\right)$
$= r \left(C o s \theta + i \sin \theta\right) \cdot s \left(C o s \phi + i \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi + i \sin \theta \cos \phi + i \cos \theta \sin \phi + {i}^{2} \sin \theta \sin \phi\right)$
$= r s \left(\cos \theta \cos \phi - \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi + \cos \theta \sin \phi\right)$
$= r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$
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:
${\tan}^{-} 1 \left(a\right) + {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a + b}{1 - a b}\right)$

$\implies {\tan}^{-} 1 \left(\frac{3}{2}\right) + {\tan}^{-} 1 \left(\frac{7}{3}\right) = {\tan}^{-} 1 \left(\frac{\left(\frac{3}{2}\right) + \left(\frac{7}{3}\right)}{1 - \left(\frac{3}{2}\right) \left(\frac{7}{3}\right)}\right) = {\tan}^{-} 1 \left(\frac{9 + 14}{6 - 21}\right)$

$= {\tan}^{-} 1 \left(\frac{23}{- 15}\right) = {\tan}^{-} 1 \left(- \frac{23}{15}\right)$

$\implies \theta + \phi = {\tan}^{-} 1 \left(- \frac{23}{15}\right)$

$r s \left(\cos \left(\theta + \phi\right) + i \sin \left(\theta + \phi\right)\right)$

$= 2 \sqrt{13} \sqrt{58} \left(\cos \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right)\right)$

$= 2 \sqrt{754} \left(\cos \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right)\right)$

This is your final answer.

You can also do it by another method.
By firstly multiplying the complex numbers and then changing it to trigonometric form, which is much easier than this.

$\left(4 + 6 i\right) \left(3 + 7 i\right) = 12 + 28 i + 18 i + 42 {i}^{2} = 12 + 46 i - 42 = - 30 + 46 i$

Now change $- 30 + 46 i$ in trigonometric form.

Magnitude of $- 30 + 46 i = \sqrt{{\left(- 30\right)}^{2} + {\left(46\right)}^{2}} = \sqrt{900 + 2116} = \sqrt{3016} = 2 \sqrt{754}$
Angle of $- 30 + 46 i = {\tan}^{-} 1 \left(\frac{46}{-} 30\right) = {\tan}^{-} 1 \left(- \frac{23}{15}\right)$
$\implies - 30 + 46 i = 2 \sqrt{754} \left(\cos \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{23}{15}\right)\right)\right)$