# Find the square roots of a = root3 + root3*i

Shackleford
Member warned about not using the homework template
I don't recall ever doing this but maybe I have.

z2 = a = p [cos Ψ + i sin Ψ] = √3 + i*√3

p = √6
Ψ = π/4

Using the formula in the notes, z = 61/4 * exp[i*(π/4 + 2π*k)/2], k = 0, 1.

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Mentor
I don't recall ever doing this but maybe I have.

z2 = a = p [cos(psi) + i sin(psi)] = root3 + i*root3

p = root6
psi = pi/4

Using the formula in the notes, z = 61/4 * exp[i*(pi/4 + 2pi*k)/2], k = 0, 1.
Do you have a question?

Also, your problem would be much more readable if you used LaTeX or the symbols available from the Advanced Menu, under the ##\Sigma## icon; e.g., √, ψ, and π.

Shackleford
When I multiply it out, I get -a.

Mentor
When I multiply it out, I get -a.
I don't. What do you get for the two roots of ##\sqrt{6}(cos(\pi/4) + i sin(\pi/4))##?

Note that in polar form, what you're calling a is (##\sqrt{6}, \pi/4##).

Shackleford
I don't. What do you get for the two roots of ##\sqrt{6}(cos(\pi/4) + i sin(\pi/4))##?

Note that in polar form, what you're calling a is (##\sqrt{6}, \pi/4##).

That's odd. I used my TI-89 to check my work, and it gave me -a. Maybe I had a typo.

Mentor
That's odd. I used my TI-89 to check my work, and it gave me -a. Maybe I had a typo.
Possibly, or maybe it's in degree mode when it should be in radian mode.

Either way, this is an easy enough problem that a calculator shouldn't be needed. ##cos(\pi/4) = sin(\pi/4) = \frac{\sqrt{2}}{2} \approx .707##

Again, what did you get for the two roots?

Shackleford
Possibly, or maybe it's in degree mode when it should be in radian mode.

Either way, this is an easy enough problem that a calculator shouldn't be needed. ##cos(\pi/4) = sin(\pi/4) = \frac{\sqrt{2}}{2} \approx .707##

Again, what did you get for the two roots?

From the original post, z2 = {61/4*ei*9π/8, 61/4*ei*π/8}

Mentor
From the original post, z2 = {61/4*ei*9π/8, 61/4*ei*π/8}
There's your mistake. Those are the correct roots of the equation z2 = ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##, but they are values of z, not z2.

If you square each of them, you get ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##.

Shackleford
There's your mistake. Those are the correct roots of the equation z2 = ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##, but they are values of z, not z2.

If you square each of them, you get ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##.

Ah. For some reason I thought that those were the square roots, not that each of those was a square root. Thanks for the help. My brain is still on vacation. Haha.