Using the results of Example 6, find the square roots of 32j.

Data from Example 6

Find the two square roots of 2j.
First, we write 2j in polar form as 2j = 2(cos 90° + j sin 90°). To find square roots, we use the exponent 1 2. The first square root is.

To find the other square root, we add 360° to 90°. This gives us

Therefore, the two square roots of 2j are 1 + j and −1 − j. We see in Fig. 12.19 that they are on a circle of radius √2 and 180° apart.

(2j)/2 = 2/2 cos +jsin- 90 2 90 2 2(cos 45 + jsin 45) = 1 + j