Question

the complex fourth roots of 5-5sqrt3i

Answer 1

The required complex root of 5-5sqrt3i are:

$10, 10(cos\frac{\pi }{6} )+isin (\frac{\pi }{6}), 10(cos\frac{\pi }{6} )+isin (\frac{\pi }{6}) ] \ and \ z_3=[10(cos\frac{\pi }{2} )+isin (\frac{\pi }{2}) ]$

Complex numbers are square roots of a negative number. In rectangular coordinates, they are expressed as:

z = x + iy

Given the complex number

$z=5-5\sqrt{3} i$

Express in polar form $z=r(cos \theta +isin \theta)$

Get the modulus

$r=|z|=\sqrt{5^2+(-5\sqrt{3})^2)} \\r=|z|=\sqrt{25+(75)} \\r=|z| = \sqrt{100}\\r=10$

Get the argument

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-5\sqrt{3}}{5}\\\theta = tan^{-1} - \sqrt{3}\\\theta = -60^0$

Since tan is negative in the second and fourth quadrant, then:

$\theta= 180 - 60\\\theta = 120^0$

The complex number in the polar form will be:

$z=10cos(cos \frac{2 \pi}{3}+isin\frac{2\pi}{3} )$

Get the fourth root of the result

$z^{\frac{1}{4} }=[10(cos(\frac{2\pi}{3} )+isin\frac{2\pi}{3} ]^{\frac{1}{4} }$

According to De Moivre's theorem:

$z^{\frac{1}{4} }=[10(cos\frac{1}{4} (\frac{2\pi}{3} )+isin\frac{1}{4} (\frac{2\pi}{3}) ]\\z^{\frac{1}{4} }=[10(cos\frac{2\pi}{12} )+isin (\frac{2\pi}{12}) ]\\z^{\frac{1}{4} }=[10(cos\frac{\pi}{6} )+isin (\frac{ \pi}{6}) ]$

Get the four roots of the complex number

$z^{\frac{1}{4} }_k=[10(cos\frac{\pi k}{6} )+isin (\frac{\pi k}{6}) ]$

If k = 0

$z_0=[10(cos0+isin 0 ]\\z_0=10$

If k = 1

$z_1=[10(cos\frac{\pi }{6} )+isin (\frac{\pi }{6}) ]$

If k = 2

$z_2=[10(cos\frac{2\pi }{6} )+isin (\frac{2\pi }{6}) ]\\z_2=[10(cos\frac{\pi }{3} )+isin (\frac{\pi }{3}) ]$

If k = 3

$z_3=[10(cos\frac{3\pi }{6} )+isin (\frac{3\pi }{6}) ]\\z_3=[10(cos\frac{\pi }{2} )+isin (\frac{\pi }{2}) ]$

From the solution above, the four roots of the complex numbers are:

$10, 10(cos\frac{\pi }{6} )+isin (\frac{\pi }{6}), 10(cos\frac{\pi }{6} )+isin (\frac{\pi }{6}) ] \ and \ z_3=[10(cos\frac{\pi }{2} )+isin (\frac{\pi }{2}) ]$

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