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

1. Jan 20, 2015

### 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.

Last edited: Jan 20, 2015
2. Jan 20, 2015

### Staff: Mentor

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.

3. Jan 20, 2015

### Shackleford

When I multiply it out, I get -a.

4. Jan 21, 2015

### Staff: Mentor

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 ($\sqrt{6}, \pi/4$).

5. Jan 21, 2015

### Shackleford

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

6. Jan 21, 2015

### Staff: Mentor

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?

7. Jan 21, 2015

### Shackleford

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

8. Jan 21, 2015

### Staff: Mentor

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 $\sqrt{6}(cos (\pi/4) + i sin (\pi/4))$.

9. Jan 21, 2015

### Shackleford

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.