Fourth Roots of -16i

By Euler's formula, $r(\cos a + i \sin a) = re^{ia}$ , where $a$ is an angle in radians. By representing $-16i$ in polar and exponential form, we obtain: $-16i = 16(\cos(3\pi/2) + i \sin(3\pi/2)) = 16e^{i\times 3\pi/2}$ . Note that since $e^{i\times 2\pi} = 1$ , we can easily rewrite $16e^{i\times 3\pi/2} = 16e^{i\times 7\pi/2} = 16e^{i\times 11\pi/2} =$ $16e^{i\times 15\pi/2} = 16e^{i\times 19\pi/2} = \ldots$ . Finding the fourth root of these equivalent numbers, we obtain the following: $2e^{i\times 3\pi/8}, 2e^{i\times 7\pi/8}, 2e^{i\times 11\pi/8}, 2e^{i\times 15\pi/8}, 2e^{i\times 19\pi/8}, \dots$ . We notice that $2e^{i\times 19\pi/8} = 2e^{i\times 3\pi/8}$ , and that the continued answers will keep on cycling within these four answers. Therefore converting these 4 roots to polar form, we have the angles equal to $3\pi/8, 7\pi/8, 11\pi/8, 15\pi/8$ , which sum to 810 degrees.

1: change -16i to euler form: 2(cos(270/4+90k)+sin(270/4+90k)), k=0, 1, 2, 3 2: plug in the values of k and get the values of theta 3: finally add the values and get the ans. 4: start smiling

The angle associated with -16i in the Real-Imaginary Plane is 270 degrees. Note that the roots are 360/4 = 90 degrees apart. The four angles {θi}, i = 1, 2, 3, 4 of the four roots are in the form 270/4 + 90 (i - 1). Hence the sum of the angles is 270 + 90 (1+2+3) = 270 + 540 = 810

As the problem stated, there are four complex root to -16i so we are dealing with the equation z^4=-16i. Now, we express -16i in polar form so that it becomes 16(cos(270)+isin(270)). Since we are dealing with roots, we can change this expression into exponential form. Using an application of de Moirve's formula, we find that this expression is equivalent to e^(i(270+360k)), where the theta mentioned above in the four roots is just 270+360k. Therefore, we can find the 4 thetas by first dividing 270+360k by 4 and then substituting in 0,1,2, and 3 for k. So the answer is 67.5+157.5+247.5+337.5=810

Let A = -16i so r = 16. I think i don't have to demonstrate a lot since this is elementary for finding such r. so A = 16( 0\pm i ) this equivalent to 16( cos \frac{3\pi}{2} + isin \frac{3\pi}{2} ) and θ= \frac{3\pi}{2} Formula : P {\frac{\3pi}{2} + 2kpi}{4} and k = 0,1,2,...,k-1

and this problem we need k = 4 so there are four cases to examine case: k=0 we have P = \frac{\3pi}{8} = {\frac{135}{2}^\circ} = θ1 case: k=1 we have P = \frac{\7pi}{8} = {\frac{315}{2}^\circ} = θ2 case: k=2 we have P = \frac{\11pi}{8} = {\frac{495}{2}^\circ} = θ3 case: k=3 we have P = \frac{\15pi}{8} = {\frac{675}{2}^\circ} = θ4

θ1+θ2+θ3+θ4 = {810^\circ}

Converting to polar coordinates, we have $|-16i| = \left(\sqrt{16^2}\right) =16$ , $-16i = 16\left(0 - i\right) = 16\left(\cos 270^\circ + i\sin 270^\circ \right)$ .

Using de Moivre's formula, we can calculate that the fourth roots are $\begin{aligned} (-16i)^{\frac{1}{4}} &= \left(16\left(\cos 270^\circ + i\sin 270^\circ \right)\right)^{\frac{1}{4}} \\ &= 16^{\frac{1}{4}}\left(\cos \left(\frac{270^\circ}{4} + \frac{360^\circ k'}{4}\right) + i\sin\left(\frac{270^\circ}{4} + \frac{360^\circ k'}{4}\right)\right) \\ &= 2\left(\cos \left(67.5^\circ + 90^\circ k'\right) + i\sin\left(67.5^\circ + 90^\circ k'\right)\right) \\ \end{aligned}$

where $k'$ is an integer. Thus, the roots are

$k' = 0$ : $z_1 = 2\left(\cos 67.5^\circ + i\sin 67.5^\circ \right)$

$k' = 1$ : $z_2 = 2\left(\cos 157.5^\circ + i\sin 157.5^\circ \right)$

$k' = 2$ : $z_3 = 2\left(\cos 247.5^\circ + i\sin 247.5^\circ \right)$

$k' = 3$ : $z_4 = 2\left(\cos 337.5^\circ + i\sin 337.5^\circ \right)$

Hence $\theta_1 + \theta_2 + \theta_3 + \theta_4 = 67.5^\circ + 157.5^\circ + 247.5^\circ + 337.5^\circ = 810^\circ$ .