Calculus Complex Number Problem Confusing

Let w = R e^αi be any nonzero complex number. Then w has n nth roots:
n solutions to the equation z^n = R e^iα or r^n * e^inθ = R e^iα
which have r = R^(1/n) and θ = α/n + 2πk/n


The nth roots of w are w^(1/n) = z = R^(1/n) e^i(α/n + 2π k/n), where k = 0, 1, 2, , n−1.

The n nth roots of w are equispaced on the circle C of radius R^(1/n) centered at the origin.

Question is...
Draw the circle C, and then find the three 3rd roots of i. The circle C has radius______
The angle difference (in radians) between adjacent 3th roots is ______

Draw the circle C, and then find the four 4th roots of -16. The circle C has radius ______
The angle difference (in radians) between adjacent 4th roots is ______

Draw the circle C, and then find the two square roots (2nd roots!) of 1 + i.
The circle C has radius ______ . The angle difference (in radians)
between adjacent square roots is ______

is this easier than it looks? and either way how do i do this?

It's exactly as easy as it looks. :^)

In each case, the radius of the circle is the nth root of r. The argument angle of the principle root is simply (Arg w)/n. And, finally, the angle between two adjacent points out of n equally-spaced points on a circle is 2π/n. That answers all of your questions, once you put w into "complex polar" form;

w = r e^(iα)

...where r = |w| and α = Arg w.

You have practically typed all of that in your first paragraph.

So, for i^(1/3):

n=3
w = i = e^(i π/2)
w = r e^(iα) form has r=1 and α=π/2.

Plug those into the formulas you typed:

Radius of circle C is r^(1/n) = 1^(1/3) = 1
Angle between adjacent roots is 2π/n = 2π/3

The other two problems are essentially the same, except that r is not 1 for those problems.

-16 = 16 e^(iπ)
1 + i = (√2) e^(i 3π/4)

It is if it is paraphrased.
w = a +bi = R(cosθ +i*sinθ) = R*e^(iθ). the nth roots of w are all z where z^n = w, so z = R^(1/n) * e^i(θ/n + 2πk/n) where k is any integer mod n.


Determine R and θ,
Plot w on the graph and draw the straight line 0 to w.
and then r=R^(1/n) and θ/n
Draw the circle radius r^(1/n) centered at 0.
Mark the point on the circle that is at θ/n.
Mark the (n-1) points that are 2π/n around from this and each subsequent point. These points form a regular n-gon on the circle. Thus...

³√i - this will be the points of an equilateral triangle,
r=1, θ=π/2 (draw a line from the origin to here )
draw circle radius ³√r = 1
mark the point (π/2)/3 = π/6 on the circle.
Then mark the two points 2π/3 and 4π/3 around from this point.

Point - line - circle - new points - n-gon