Help converting complex number to standard form!?

It's best to use the form r(cos(θ) + isin(θ)) for a problem like this as it allows you to use De Moivre's theorem. Given a + bi:

r = √(a² + b²)

θ = tan⁻¹(b/a)

In this example:

r = 2

θ =π/6

Now you would want to verify that the angle is correct, which in this case it is since √3 + i is in the 1st quadrant. But you'd want to verify where the angle is so that you use the correct angle for De Moivre's theorem. De Moivre's theorem says:

(r(cos(θ) + isin(θ)))ⁿ = rⁿ(cos(nθ) + isin(nθ))

Rewrite √3 + i:

2(cos(π/6) + i sin(π/6))

Now raise to 100

(2(cos(π/6) + i sin(π/6)))¹⁰⁰ = 2¹⁰⁰(cos(100π/6) + isin(100π/6))

Convert 100π/6 to and angle on [0,2π).

100π/6 = 16π + 4π/6 = 16π + 2π/3

16π is just 8 full rotations, and your left with 2π/3, so (√3 + i)¹⁰⁰ turns into:

2¹⁰⁰(cos(2π/3) + isin(2π/3))

Which reduces to:

2¹⁰⁰(-¹/₂ + i√3/2)

Now distribute to get:

-2⁹⁹ + i2⁹⁹√3

That's a big number, about -0.6338x10^30 + 1.0978x10^30 i.

In order to compute it, first convert (sqrt(3) + i ) to polar form, r @ angle

Then (r @ angle)^100 is r^100 @ (100 angle).

r^100 is not so hard to figure, and for 100 angle you need to subtract multiples of 2pi (or 360 degrees).

Once you have done all that, convert back to standard xy form.

( √3 + i ) = 2 [ cos π/6 + i sin π / 6 ] = 2 e^(iπ/6) --->

answer is 2^(100) [ cos (100π/6 ) + i sin (100π / 6 ) ] = 2^100 [ - 1/2 + i √3 / 2 ]