Complex Angles

Draw the point 4-4sqrt(3)i on a diagram.

Use trig to see it makes an angle of 30 degrees with the negative y axis.

It's argument is therefore 270+30=300

Divide that by 4 to get the argument of one of the roots: 75

The 4th roots differ by 90 degrees, check that these are the arguments of a complete set of roots:

75 75+90 75+180 75+270

Then add them up to get the answer.

It is clear that $z=4-4\sqrt3 i=8(\cos 300º + i \sin 300º)$ , so the first root is $\sqrt[4]{8}(\cos 75º + i \sin 75º)$ and the other ones are obtained by adding 90º to the argument. The desired sum is $75+(75+90)+(75+2\cdot 90)+(75+3\cdot90)=840$

Let $x = 4-4\sqrt{3}i$

First, convert $x$ into polar form.

Magnitude = $|x|=\sqrt{4^2+(4\sqrt{3})^2} = 8$

Angle = $\tan^{-1} (\frac{-4\sqrt{3}}{4}) = \tan^{-1}(-\sqrt{3}) = -60^\circ +180^\circ k$ , where $k$ is any integer. When graphed, $x$ lies in the fourth quadrant, therefore we know that the angle must be in the fourth quadrant. Thus the angle is $-60^\circ+360^\circ k$ , and cannot be $120^\circ+360^\circ k$ , the $\tan$ of which also happens to be $-\sqrt{3}$ .

Thus $x = 8(\cos(-60^\circ+360^\circ k) + i\sin(-60^\circ+360^\circ k)) = 8cis(-60^\circ+360^\circ k)$

To find the fourth roots of $x$ , we raise $x$ to the $\frac{1}{4}$ power. By De Moivre's theorem, we know that when taking a complex number in polar form to a power $n$ , we raise the magnitude to the $n$ th power, and we multiply the angle by $n$ .

Thus we have:

$x^{\frac{1}{4}} = 8^{\frac{1}{4}}cis(\frac{-60^\circ+360^\circ k}{4}) = 8^{\frac{1}{4}}cis(-15^\circ+90^\circ k)$

We are only concerned with the sum of the angles, so all we must look for are the values of $k$ for which $0^\circ \leq -15^\circ+90^\circ k \leq 360^\circ$

It is easy to see that the four $k$ s we need are 1, 2, 3, and 4. Thus our four angles are $75^\circ, 165^\circ, 255^\circ, 345^\circ$ . The sum of these is 840.

The number $4-4\sqrt{3}i$ can be rewritten as $8(\frac{1}{2}-\frac{\sqrt{3}}{2}i) = r(\cos\theta + i\sin\theta).$

Therefore, for $\theta$ in the range 0 to 360, $r=8, \space\cos\theta = \frac{1}{2},\space \sin\theta = -\frac{\sqrt{3}}{2} \Longrightarrow \theta=300^\circ$ .

By DeMoivre's theorem, we see that: $\sqrt[4]{4-4\sqrt{3}i} = \sqrt[4]{8}(\cos\frac{\theta + 360^\circ k}{n} + i\sin\frac{\theta+360^\circ k}{n})$ for $k \in \{0,1,2,3\}$ .

Substituting $300^\circ$ for $\theta$ , $4$ for $n$ , and $\{0,1,2,3\}$ for $k$ , we get $\theta_1+\theta_2+\theta_3+\theta_4 = 75^\circ + 165^\circ + 255^\circ + 345^\circ = \boxed{840^\circ}.$

First we see that $4-4\sqrt{3}i=8(\cos (\frac{-\pi}{3}+k2\pi ) +i \sin (\frac{-\pi}{3}+ k2\pi ))$ So i'ts easy to conclude that four complex fourth roots are $z_k=\sqrt[4]{8}(\cos (\frac{-\pi}{12}+\frac{k\pi}{2})+ i\sin (\frac{-\pi}{12}+\frac{k\pi}{2}) , k=1,2,3,4$ ( because $0^o \leq \theta_k \leq 360^o$

So $\theta_1+\theta_2+\theta_3+\theta_4= -4.\frac{\pi}{12}+ \frac{\pi(1+2+3+4)}{2} =\frac{14\pi}{3}$

Or $\theta_1+\theta_2+\theta_3+\theta_4= \frac{14}{3}.180^o= 840^o$ degrees

Let define cis \theta = \cos \theta + i\times \sin \theta

So, \sqrt[4]{4 - 4\sqrt{3}} = \sqrt[4]{8(\frac {1}{2} - \frac {\sqrt{3}}{2}} = = \sqrt[4]{8} \times + sqrt[4]{cis (300^\circ + n.360^\circ}.

But (cis \theta) ^ \alpha = (cis \theta \times\alpha)

So, using only the angular part, we have:

sqrt[4]{cis (300^\circ + n.360^\circ} = (cis (300^\circ + n.360^\circ))^(\frac {1}{4 }) = (cis (300^\circ \times (\frac {1}{4 })+ n.360^\circ) \times (\frac {1}{4 })) = {cis (75^\circ + n.90^\circ}

For n = 0, the first angle is 75^\circ For n = 1, the second angle is 75^\circ + 90^\circ = 165^\circ For n = 2, the third angle is 75^\circ + 180^\circ = 255^\circ For n = 3, the fouth angle is 75^\circ + 270^\circ = 345^\circ

and the sum os these angles are: 75^\circ +165^\circ + 255^\circ + 345^\circ = 840^\circ.

Euler's Formula states that e^{i\theta} = \cos(\theta) + i \sin(\theta) We can use this to find roots of complex numbers. The nth roots of a complex number in the form e^{i\theta} are e^{\frac{\theta + 2k\pi}{n}} where k is 0,1,2,\dots$n-1. 4-4\sqrt{3}i can be rewritten as 8(\frac{1}{2} - \frac{\sqrt{3}}{2}i) \rightarrow 8e^{i300^{\circ}}. The 4th roots are thus \sqrt[4]{8}e^{i75^{\circ}}, \sqrt[4]{8}e^{i165^{\circ}}, \sqrt[4]{8}e^{i255^{\circ}}, \sqrt[4]{8}e^{i345^{\circ}}. 75+165+255+345 = 840

By plotting the complex number $z = 4 - 4\sqrt{3}$ we see that it rotates in a clockwise direction from the positive x axis.

This clockwise rotation is given by the negative argument $arg(z) \equiv -\tan^{-1} \left( \frac{4 \sqrt{3} }{4} \right) = -\tan^{-1}(\sqrt{3}) = -60^\circ$ .

Converting this to a positive (anti-clockwise) argument, we get $arg (z) = 360 - 60 = 300^\circ$ .

As we are looking for the fourth roots of this complex number, any solution $z_i$ must satisfy De Moivre's formula, specifically: $z = z_i^{4} = r^{4} [ \cos (4 \theta_i) + i\sin (4 \theta_i) ]$

(This can be seen be rewriting the complex numbers in exponential polar form, ie. $re^{i \theta} = r(\cos \theta + i \sin \theta )$ and applying index laws).

Also, as complex roots are multi-valued, each of these arguments $(4 \theta_i)$ need only be congruent to $300^\circ \pmod{360}$ :

$4 \theta_i \equiv 300 \pmod {360} \\ 4 \theta_i \equiv 300 + k \times 360 \text{ where } k \in \mathbb{Z}$ .

However as we only want solutions where $0^\circ \leq \theta_i < 360^\circ$ we will only need to look at the 4 solutions given by $k \in \{0,1,2,3\}$ .

Therefore $4 \theta_i = 300^\circ, 660^\circ, 1020^\circ, 1380^\circ \\ \theta_i = 75^\circ, 165^\circ, 255^\circ, 345^\circ \\ \sum_{i=1}^4 \theta_i = 75+165+255+345 = 840$

$4(1-\sqrt{3}) = 8 \times (\frac{1}{2}-\frac{\sqrt{3}}{2}) = 8 \times e^{i \times (2k\pi+\frac{5\pi}{3})} = 8 \times cis(2k\pi+\frac{5\pi}{3})$

$(8 \times cis(2k\pi+\frac{5\pi}{3}))^{\frac{1}{n}} = 8^{\frac{1}{n}} \times cis(\frac{2k\pi+\frac{5\pi}{3}}{n})$

Here, $n = 4$

For $0 <= \theta_{i} <= 2\pi,$ we have,

$0 <= \frac{2k\pi+\frac{5\pi}{3}}{4} <= 2\pi$

Hence, $0 <= 2k+\frac{5}{3} <= 8$

$0 <= k <= 19/6$

Therefore, $k = 0, 1, 2, 3$

$\theta_{1} = \frac{\frac{5\pi}{3}}{4}, \theta_{2} = \frac{2\pi+\frac{5\pi}{3}}{4}, \theta_{3} = \frac{4\pi+\frac{5\pi}{3}}{4}, \theta_{4} = \frac{6\pi+\frac{5\pi}{3}}{4}$

$\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4} = (3+\frac{5}{3})\pi = \frac{14\pi}{3} = 840$

The number $z = 4 - 4\sqrt{3}i$ can be written in polar form as $z = 8 (\frac{1}{2} + -\frac{\sqrt{3}}{2})$ , thus giving us the equation for its argument as

\begin{align} \cos{\theta} = \frac{1}{2}, \; \; \; \sin{\theta} = -\frac{\sqrt{3}}{2} \end{align}

Clearly $\theta = 300^{\circ}$ .

The arguments of $\sqrt[4]{z}$ can be written as $\frac{300^{\circ} + 360n^{\circ}}{4} = 75^{\circ} + 90n^{\circ}$ , from $n = 0$ to $3$ .

Thus, the sum of all of them equals $4 \times 75^{\circ} + 6 \times 90^{\circ} = \boxed{840^{\circ}.}$

First, we convert $4 - 4\sqrt3i$ to the exponential form $\rightarrow$ $8e^{\frac{5}{3}i\pi}$

$\frac{5}{3}\pi$ is equivalent to $\frac{11}{3}\pi$ , $\frac{17}{3}\pi$ and $\frac{23}{3}\pi$ .

The fourth root of $e^{\frac{5}{3}i\pi} = (e^{\frac{5}{3}i\pi})^{\frac{1}{4}} = e^{\frac{5}{12}i\pi}$ , and using the same method, the fourth roots of $\frac{11}{3}\pi$ , $\frac{17}{3}\pi$ and $\frac{23}{3}\pi$ are $e^{\frac{11}{12}i\pi}$ , $e^{\frac{17}{12}i\pi}$ and $e^{\frac{23}{12}i\pi}$ .

Converting the radians to degrees, the angles are 75, 165, 255 and 345. (there are more equivalent but these are the only four between 0 and 360).

$75 + 165 + 255 + 345 = \boxed{840}$

Here, we need to calculate $(4 - 4\sqrt{3}i)^{\frac{1}{4}}$

Write the complex number in polar form, i.e. writing $a + ib$ in the form $re^{i\theta}$ where $r = \sqrt{a^{2} + b^{2}}$ and $\theta = \tan{\frac{b}{a}}$

Hence, we need to calculate $(8e^{-\pi/3})^{\frac{1}{4}}$

Here, since only the angles are needed to be evaluated, we consider only the exponent term.

So, we have $(e^{-\pi/3})^{\frac{1}{4}}$

Since it has $4$ complex values, we multiply the term inside the bracket with $e^{2\pi k}$ as its value is $1$ , where $k$ goes from $0$ to $(4 - 1)$ , i.e. $0$ to $3$ .

So, we have $(e^{-\pi/3} \times e^{2\pi k})^{\frac{1}{4}}$

$(e^{-\pi/3 + 2\pi k})^{\frac{1}{4}}$

Put different values of $k$ from $0$ to $3$ , we get

$\theta 1 = \frac{-\pi}{12}$

$\theta 2 = \frac{5\pi}{12}$

$\theta 3 = \frac{11\pi}{12}$

$\theta 4 = \frac{17\pi}{12}$

It is given in the question that $\theta$ lies between $0^{o}$ to $360^{o}$ , i.e. $0$ radians to $2\pi$ radians.

Hence, we need to modify $\theta 1$ . $\frac{-\pi}{12}$ is equivalent to $2\pi - \frac{\pi}{12} = \frac{23\pi}{12}$ . So, $\theta 1 = \frac{23\pi}{12}$

Adding all the $4$ $\theta$ 's, we get the value

$\theta 1 + \theta 2 + \theta 3 + \theta 4 = \frac{56\pi}{12}$ radians $= \frac{56 \times 180}{12}$ degrees.

$= 840$ degrees

That's the answer!