# Write the complex numbers 4cis120^@ and 9cis((3pi)/2) in the form a+ib?

May 31, 2017

See explanation.

#### Explanation:

I assume that the "cis" is an abbreviation of $\left(\cos + i \sin\right)$. If we use the formula we get:

$B i$

$4 \cdot \left(\cos 120 + i \sin 120\right) = 4 \cdot \left(\cos \left(90 + 30\right) + i \sin \left(90 + 30\right)\right)$

$= 4 \cdot \left(- \sin 30 + i \cos 30\right) = 4 \cdot \left(- \frac{1}{2} + i \cdot \frac{\sqrt{3}}{2}\right) = - 2 + 2 \sqrt{3} i$

As real component is negative and imaginary component is positive, this lies in second quadrant.

$B i i$

$9 \cdot \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right) = 9 \cdot \left(0 + i \cdot \left(- 1\right)\right) = - 9 i$

As real component is zero and imaginary component is negative, this lies on imaginary axis, $9$ points below $0 + i 0$.

The sketched numbers are as shown below:

graph{(x^2+(y+9)^2-0.12)((x+2)^2+(y-2sqrt3)^2-0.1)=0 [-20.5, 19.5, -12.44, 7.56]}