complex numbers?

Question

express : squareroot of 3 + i
in polar form

ah the answers are different
who is right?

Answer 1

To take the square root of a complex number, you place it in polar coordinates, take the square root of the magnitude (r) and half of the direction (theta).

So the first thing to do would be to put 3 + i into polar form.
r = sqrt(3^2 + 1^2) = sqrt(10) = 3.162278
theta = tan^-1(1/3) = 0.32175

Now, that's not the square root of 3 + i, that's just 3 + i.

Next, to take the square root. r of the square root is the square root of the square root of 10, which is the fourth root of 10 (about 1.77828). Theta of the square root is half of 0.32175, which is about 0.160875

That's the answer:
r = 1.77828
theta = 0.160875 (radians)

Answer 2

I think the other people took your problem to be either 3 + i (forgetting the square root) or the square root of the entire complex number. I read it as only the 3 is in the square root, which means I'll have a third answer for you!

If I graph sqr(3) + i on the plane and then draw the vector from (0,0) to the point, I see a right triangle with legs sqr(3) on the x-axis and 1 on the y-axis.

From there I can use the pythagorean theorem to get the length of the hypotenuse, which is the r of the polar form.

And it looks like a 30-60-90 triangle to me!

Answer 3

r = √3 + i
|r| = √(3 + 1) = √4 = 2
tan θ = 1 / √3
θ = 30°
r = 2 (cos 30° + i sin 30°)
r = 2 /_30°

Answer 4

3 + i = r*e^(i*θ) = √10 e^[i*arctan(1/3)]
where r = √(1+3^2) = √10, θ = arctan(1/3)