Find two square roots for the following complex number. Leave your answers in trigonometric form.25(cos 290° + i sin 290°)

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:

r^(1/n)[cos((θ+2πk)/n) + i sin((θ+2πk)/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 2 for square root), and k is an integer from 0 to n-1.

Given the complex number 25(cos 290° + i sin 290°), we have:

r = 25 θ = 290° n = 2

We need to find two square roots, so we will use k = 0 and k = 1.

For k = 0:

root1 = 25^(1/2)[cos((290°+2π0)/2) + i sin((290°+2π0)/2)] root1 = 5[cos(145°) + i sin(145°)]

For k = 1:

root2 = 25^(1/2)[cos((290°+2π1)/2) + i sin((290°+2π1)/2)] root2 = 5[cos(215°) + i sin(215°)]

So, the two square roots of the complex number 25(cos 290° + i sin 290°) in trigonometric form are 5[cos(145°) + i sin(145°)] and 5[cos(215°) + i sin(215°)].

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:

r^(1/n)[cos((θ+2kπ)/n) + i sin((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 2 for square root), and k is an integer from 0 to n-1.

Given the complex number 25(cos 290° + i sin 290°), we have:

r = 25 θ = 290° n = 2

We can find the two square roots by plugging these values into the formula for k=0 and k=1.

For k=0:

25^(1/2)[cos((290°+20π)/2) + i sin((290°+20π)/2)] = 5[cos(145°) + i sin(145°)]

For k=1:

25^(1/2)[cos((290°+21π)/2) + i sin((290°+21π)/2)] = 5[cos(215°) + i sin(215°)]

So, the two square roots of the complex number 25(cos 290° + i sin 290°) are 5[cos(145°) + i sin(145°)] and 5[cos(215°) + i sin(215°)].