# How can you use trigonometric functions to simplify  16 e^( ( 7 pi)/6 i )  into a non-exponential complex number?

Jul 29, 2016

I used the exponential converion relationship:

#### Explanation:

You may remember that your complex number can be written as:
$z = \rho {e}^{\theta i} = \rho \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$
$z = 16 {e}^{\frac{7}{6} \pi i} = 16 \left[\cos \left(\frac{7}{6} \pi\right) + i \sin \left(\frac{7}{6} \pi\right)\right]$
we can now evaluate the above trig functions and multiply by $16$ to get to the form: $z = a + i b$:
$z = - 13.8 - 8 i$