Find two square roots for the following complex number. Leave your answers in trigonometric form.4(cos 40° + i sin 40°)

To find the square roots of a complex number in trigonometric form, we use De Moivre's Theorem. The formula for the nth root of a complex number in trigonometric form is given by:

r^(1/n)[cos((θ+2kπ)/n) + isin((θ+2kπ)/n)]

where:

• r is the modulus of the complex number
• θ is the argument of the complex number
• n is the root you want to find (in this case, n=2 because we're finding the square root)
• k is an integer that varies from 0 to n-1

Given the complex number 4(cos 40° + i sin 40°), we have r=4 and θ=40°.

Step 1: Find r^(1/n) The modulus of the square roots is the square root of the modulus of the original complex number, so we have r^(1/n) = 4^(1/2) = 2.

Step 2: Find the arguments of the square roots The arguments of the square roots are given by (θ+2kπ)/n for k=0 and k=1.

For k=0: (40°+20π)/2 = 20° For k=1: (40°+21π)/2 = 200°

Step 3: Write the square roots in trigonometric form So the two square roots of the complex number 4(cos 40° + i sin 40°) are:

2(cos 20° + i sin 20°) and 2(cos 200° + i sin 200°)