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

[Shackleford]

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 = 6^1/4 * exp[i*(π/4 + 2π*k)/2], k = 0, 1.

 

[Mark44]

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 = 6^1/4 * 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.

 

[Mark44]

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.

 

[Mark44]

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 = {6^1/4*e^i*9π/8, 6^1/4*e^i*π/8}

 

[Mark44]

From the original post, z² = {6^1/4*e^i*9π/8, 6^1/4*e^i*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.