# How do you divide  (2+i)/(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 $\left(2 + i\right)$ and $\theta$ be its angle.
Magnitude of $\left(2 + i\right) = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{4 + 1} = \sqrt{5} = r$
Angle of $\left(2 + i\right) = T a {n}^{-} 1 \left(\frac{1}{2}\right) = {\tan}^{-} 1 \left(\frac{1}{2}\right) = \theta$

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

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,
$\frac{2 + i}{3 + 7 i}$

$= \frac{r \left(C o s \theta + i \sin \theta\right)}{s \left(C o s \phi + i \sin \phi\right)}$

=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi

$= \frac{r}{s} \cdot \frac{\cos \theta \cos \phi + i \sin \theta \cos \phi - i \cos \theta \sin \phi - {i}^{2} \sin \theta \sin \phi}{{\cos}^{2} \phi - {i}^{2} {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi - \cos \theta \sin \phi\right)}{{\cos}^{2} \phi + {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)}{1}$

$= \frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

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 $\theta - \phi$.

$\theta - \phi = {\tan}^{-} 1 \left(\frac{1}{2}\right) - {\tan}^{-} 1 \left(\frac{7}{3}\right)$
We know that:
${\tan}^{-} 1 \left(a\right) - {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a - b}{1 + a b}\right)$

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

$= {\tan}^{-} 1 \left(\frac{3 - 14}{6 + 7}\right) = {\tan}^{-} 1 \left(- \frac{11}{13}\right)$

$\implies \theta - \phi = {\tan}^{-} 1 \left(- \frac{11}{13}\right)$

$\frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

$= \frac{\sqrt{5}}{\sqrt{58}} \left(\cos \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right)\right)$

$= \sqrt{\frac{5}{58}} \left(\cos \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right)\right)$

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

First of all let's simplify the given number
$\frac{2 + i}{3 + 7 i}$.

Multiply and divide by the conjugate of the complex number present in the denominator i.e $3 - 7 i$.

$\frac{2 + i}{3 + 7 i} = \frac{\left(2 + i\right) \left(3 - 7 i\right)}{\left(3 + 7 i\right) \left(3 - 7 i\right)} = \frac{6 - 14 i + 3 i - 7 {i}^{2}}{{3}^{2} - {\left(7 i\right)}^{2}}$
$= \frac{6 - 11 i + 7}{9 - \left(- 49\right)} = \frac{13 - 11 i}{9 + 49} = \frac{13 - 11 i}{58} = \frac{13}{58} - \frac{11 i}{58}$

$\frac{2 + i}{3 + 7 i} = \frac{13}{58} - \frac{11 i}{58}$

Let $t$ be the magnitude of $\left(\frac{13}{58} - \frac{11 i}{58}\right)$ and $\beta$ be its angle.
Magnitude of $\left(\frac{13}{58} - \frac{11 i}{58}\right) = \sqrt{{\left(\frac{13}{58}\right)}^{2} + {\left(- \frac{11}{58}\right)}^{2}} = \sqrt{\frac{169}{3364} + \frac{121}{3364}} = \sqrt{\frac{290}{3364}} = \sqrt{\frac{5}{58}} = t$
Angle of $\left(\frac{13}{58} - \frac{11 i}{58}\right) = T a {n}^{-} 1 \left(\frac{- \frac{11}{58}}{\frac{13}{58}}\right) = {\tan}^{-} 1 \left(- \frac{11}{13}\right) = \beta$

$\implies \left(\frac{13}{58} - \frac{11 i}{58}\right) = t \left(C o s \beta + i \sin \beta\right)$
$\implies \left(\frac{13}{58} - \frac{11 i}{58}\right) = \sqrt{\frac{5}{58}} \left(C o s \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{11}{13}\right)\right)\right)$.