Fourth roots of -8 + 8i√3

[calc1]

I understand how to get cube roots of -i, 64, or -16 for example, but. . .

using: nth root of z = (nth root of r)*[cos((x+2πk)/n) + i * sin((x+2πk)/n)], where k = 0, 1, 2, . . ., n-1, x = angle of z. I hope you can understand which formula I am using,

How do I find the fourth roots of -8 + 8i√3? Would I start with a division by 8 then graph at (-1, √3)? I get angle 3π/2, but I using the nth root of the radius as the first exponent is not working out for me.

We are in a chapter using DeMoivre's Theorem to raise a complex number to a power, but are working on roots.

 

[Deleted member 4993]

I understand how to get cube roots of -i, 64, or -16 for example, but. . .

using: nth root of z = (nth root of r)*[cos((x+2πk)/n) + i * sin((x+2πk)/n)], where k = 0, 1, 2, . . ., n-1, x = angle of z. I hope you can understand which formula I am using,

How do I find the fourth roots of -8 + 8i√3? Would I start with a division by 8 then graph at (-1, √3)? I get angle 3π/2, but I using the nth root of the radius as the first exponent is not working out for me.

We are in a chapter using DeMoivre's Theorem to raise a complex number to a power, but are working on roots.

-8 + 8i√3 = 16*[-1/2 + i (√3)/2]

 

[calc1]

-8 + 8i√3 = 16*[1/2 - i (√3)/2]

Thanks for the absolute value. That was a million miles ago last week.

2[cos((angle? + 2πk)/4) + i*sin((angle? + 2πk)/4)]

My final answers are the four variations of ±√3±i, but I do not know what coordinates to use for my initial angle.

 

[HallsofIvy]

The modulus (absolute value) of \(\displaystyle -8+ 8i\sqrt{3}\) is \(\displaystyle \sqrt{(-8)^2+ (8\sqrt{3})^2}= \sqrt{64+ 192}= \sqrt{256}= 16\).

The argument (angle) is \(\displaystyle arctan\left(\frac{8\sqrt{3}}{8}\right)= arctan(\sqrt{3})\). The fourth roots have modulus the fourth root of 16 and argument \(\displaystyle arctan(\sqrt{3})\) over 4.

(Corrected thanks to Subhotosh Khan pointing out that \(\displaystyle 8^2\) is NOT 8!)

 

[Deleted member 4993]

I understand how to get cube roots of -i, 64, or -16 for example, but. . .

using: nth root of z = (nth root of r)*[cos((x+2πk)/n) + i * sin((x+2πk)/n)], where k = 0, 1, 2, . . ., n-1, x = angle of z. I hope you can understand which formula I am using,

How do I find the fourth roots of -8 + 8i√3? Would I start with a division by 8 then graph at (-1, √3)? I get angle 3π/2, but I using the nth root of the radius as the first exponent is not working out for me.

We are in a chapter using DeMoivre's Theorem to raise a complex number to a power, but are working on roots.

-8 + 8i√3 = 16*[cos(2π/3 + 2kπ) + i sin(2π/3 + 2kπ)] = 2⁴*e^{(2π/3 + 2kπ)}

(-8 + 8i√3)^1/4 = ±2*e^{(π/6 + kπ/2)}

 

[lookagain]

-8 + 8i√3 = 16*[1/2 - i (√3)/2]

You meant -8 + 8i√3 = -16*[1/2 - i (√3)/2]

 

[Deleted member 4993]

You meant -8 + 8i√3 = -16*[1/2 - i (√3)/2]

No I meant what I wrote later (I think)

-8 + 8i√3 = 16*[-1/2 + i (√3)/2]

cos(2π/3) = -1/2

sin(2π/3) = √(3)/2

 

[calc1]

The modulus (absolute value) of \(\displaystyle -8+ 8i\sqrt{3}\) is \(\displaystyle \sqrt{(-8)^2+ (8\sqrt{3})^2}= \sqrt{64+ 192}= \sqrt{256}= 16\).

OK.

r = modulus, n = number of roots. Then, r^(1/n)cis[(z + (360*k)/n], k = 0, 1, 2, 3.

16^(1/4) = 2

Angle z? (-8,8√3) denotes QII, then tan= sin/cos: (8√3)/-8 = -√3, arctan that for reference angle -60, then 180-60, so this angle is 120 degrees.

2cis[(120 + (360 * k))/4]

The four roots:
k = 0, 2cis(30), 2(cos(30) + i sin(30)), (2 * (√3/2)) + i(2 * (1/2)), √3 +i
k = 1, 2cis(120), -1 + i√3
k = 2, 2cis(210), -√3 - i
k = 3, 2cis(300), 1 - i√3