Convert - √3-i to polar form.

a. 2cis (pi) / 6

b. 2cis5 (pi) / 6

c. 2cis7 (pi) / 6

d. 2cis11 (pi) / 6

e. 2cis4 (pi) / 6

-sqrt (3) - i = 2 cis (7pi/6)

Step-by-step explanation:

-sqrt (3) - i

We can find the radius

r = sqrt ((-sqrt (3)) ^2 + (-1) ^2)

= sqrt (3 + 1)

= sqrt (4)

= 2

theta = arctan (y/x)

arctan (-1/-sqrt (3))

arctan (1/sqrt (3))

theta = pi/6

But this is in the first quadrant, and we need it in the third quadrant

Add pi to move it to the third quadrant

theta = pi/6 + 6pi/6

=7pi/6

-sqrt (3) - i = 2 cis (7pi/6)