Question

Find all the complex roots. Leave your answers in polar form with the argument in degrees. The complex fourth roots of -9i.

Answer 1

Answer:

Step-by-step explanation:

Given z = x+iy

The polar form of a complex number $z = r(cos\theta + isin\theta)$

The nth root of the complex number is expressed according to de moivre's  theorem as;

$z^{\frac{1}{n} } = [r(cos\theta + isin\theta)]^{\frac{1}{n} } \\z^{\frac{1}{n} } = \sqrt[n]{r} (cos(\frac{\theta+2nk}{n} ) + isin(\frac{\theta+2nk}{n}))$

r is the modulus of the complex  number and $\theta$ is the argument

r = √x²+y²

$\theta = tan^{-1} \frac{y}{x}$

Given z = -9i

r = √0+(-9)²

r = √81

r = 9

$\theta = tan^{-1}\frac{-9}{0} \\\theta = tan^{-1}-\infty\\\theta = -90^{0}$

The argument will be equivalent to 180-90 = 90°

The forth root of -9i will be expressed as shown according to de moivre's theorem;

$z_k^{\frac{1}{4} } = \sqrt[4]{9} (cos(\frac{90+2(4)k}{4} ) + isin(\frac{90+2(4)k}{4}))\\z_k^{\frac{1}{4} } = \sqrt[4]{9} (cos(\frac{90+8k}{4} ) + isin(\frac{90+8k}{4}))$

The complex roots are at when k = 0, 1, 2 and 3

When k = 0;

$z_0 = \sqrt[4]{9} (cos(\frac{90}{4} ) + isin(\frac{90}{4}))\\z_0 = \sqrt[4]{9} (cos(23) + isin(23))\\\\when\ k =1\\z_1 = \sqrt[4]{9} (cos(\frac{90+8}{4} ) + isin(\frac{90+8}{4}))\\z_1 = \sqrt[4]{9} (cos(25 ) + isin(25))\\\\when\ k =2\\z_2 = \sqrt[4]{9} (cos(\frac{90+16}{4} ) + isin(\frac{90+16}{4}))\\z_2 = \sqrt[4]{9} (cos(27 ) + isin(27))\\\\when\ k =3\\z_3 = \sqrt[4]{9} (cos(\frac{90+24}{4} ) + isin(\frac{90+24}{4}))\\z_3 = \sqrt[4]{9} (cos(29 ) + isin(29))$

Note that all the degrees are rounded to the nearest whole number.