# How do you add (2+8i)+(3+4i) in trigonometric form?

May 18, 2016

$\left(2 + 8 i\right) + \left(3 + 4 i\right) = \sqrt{68} \left[\cos \alpha + i \sin \alpha\right] + 5 \left[\cos \beta + i s \sin \beta\right]$,

where $\alpha = \arctan 4$ and $\beta = \arctan \left(\frac{4}{3}\right)$.

#### Explanation:

Let us first write $\left(2 + 8 i\right)$ and $\left(3 + 4 i\right)$ in trigonometric form.

$a + i b$ can be written in trigonometric form $r \cos \theta a + i r \sin \theta = r \left(\cos \theta a + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\tan \theta = \frac{b}{a}$ or $\theta = \arctan \left(\frac{b}{a}\right)$

Hence $\left(2 + 8 i\right) = \sqrt{{2}^{2} + {8}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{68} \left[\cos \alpha + i \sin \alpha\right]$, where $\alpha = \arctan 4$ and

$\left(3 + 4 i\right) = \sqrt{{3}^{2} + {4}^{2}} \left[\cos \beta + i \sin \beta\right]$ or

$5 \left[\cos \beta + i s \sin \beta\right]$, where $\beta = \arctan \left(\frac{4}{3}\right)$

Hence $\left(2 + 8 i\right) + \left(3 + 4 i\right)$ =

$\sqrt{68} \left[\cos \alpha + i \sin \alpha\right] + 5 \left[\cos \beta + i s \sin \beta\right]$,

where $\alpha = \arctan 4$ and $\beta = \arctan \left(\frac{4}{3}\right)$.