Answer:
Step-by-step explanation:
Given z = x+iy
The polar form of a complex number [tex]z = r(cos\theta + isin\theta)[/tex]
The nth root of the complex number is expressed according to de moivre's theorem as;
[tex]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}))[/tex]
r is the modulus of the complex number and [tex]\theta[/tex] is the argument
r = √x²+y²
[tex]\theta = tan^{-1} \frac{y}{x}[/tex]
Given z = -9i
r = √0+(-9)²
r = √81
r = 9
[tex]\theta = tan^{-1}\frac{-9}{0} \\\theta = tan^{-1}-\infty\\\theta = -90^{0}[/tex]
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;
[tex]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}))\\[/tex]
The complex roots are at when k = 0, 1, 2 and 3
When k = 0;
[tex]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))\\[/tex]
Note that all the degrees are rounded to the nearest whole number.
