Question

Find all complex cube roots of -2-i
. Give your answers in a+bi form

Answer 1

so we have a point at -2-i or -2 - 1i, that means that both "x" and "y" are negative, the only occurs in the III Quadrant, so hmmm let's find the modulus and angle θ

`\stackrel{a}{-2}\stackrel{b}{-1i}\hspace{5em} \begin{cases} r=√((-2)^2 + (-1)^2)\\ \qquad √(5)\\ \theta =tan^(-1)\left( (-1)/(-2) \right)\\[1em] \qquad \approx 206.57^o \end{cases} \\\\\\ \stackrel{\textit{let's keep in mind that}}{\sqrt[3]{√(5)}\implies \left( 5^{(1)/(2)} \right)^{(1)/(3)}}\implies 5^{(1)/(6)}\implies \sqrt[6]{5} \\\\[-0.35em] ~\dotfill`

`\sqrt[n]{z}=\sqrt[n]{r}\left[ \cos\left( \cfrac{\theta+2\pi k}{n} \right) +i\sin\left( \cfrac{\theta+2\pi k}{n} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\boxed{k=0}\hspace{5em} \sqrt[ 3 ]{√(5)} \left[ \cos\left( \cfrac{ 206.57^o + 360^o( 0 )}{3} \right) +i \sin\left( \cfrac{ 206.57^o + 360^o( 0 )}{3} \right)\right] \\\\\\ \sqrt[ 6 ]{5} \left[ \cos\left( \cfrac{ 206.57^o }{3} \right) +i \sin\left( \cfrac{ 206.57^o }{3} \right)\right] \\\\\\ \sqrt[6]{5}\left[ \cos(68.86^o) +i \sin(68.86^o)\right] ~~ \approx ~~ 0.47~~ + ~~1.22i \\\\[-0.35em] ~\dotfill`

`\boxed{k=1}\hspace{5em} \sqrt[ 6 ]{5} \left[ \cos\left( \cfrac{ 206.57^o + 360^o( 1 )}{3} \right) +i \sin\left( \cfrac{ 206.57^o + 360^o( 1 )}{3} \right)\right] \\\\\\ \sqrt[ 6 ]{5} \left[ \cos\left( \cfrac{ 566.57^o }{3} \right) +i \sin\left( \cfrac{ 566.57^o }{3} \right)\right] \\\\\\ \sqrt[ 6 ]{5} \left[ \cos(188.86^o) +i \sin(188.86^o)\right] ~~ \approx ~~ -1.29~~ - ~~0.20i \\\\[-0.35em] ~\dotfill`

`\boxed{k=2}\hspace{5em} \sqrt[ 6 ]{5} \left[ \cos\left( \cfrac{ 206.57^o + 360^o( 2 )}{3} \right) +i \sin\left( \cfrac{ 206.57^o + 360^o( 2 )}{3} \right)\right] \\\\\\ \sqrt[ 6 ]{5} \left[ \cos\left( \cfrac{ 926.57^o }{3} \right) +i \sin\left( \cfrac{ 926.57^o }{3} \right)\right] \\\\\\ \sqrt[ 6 ]{5} \left[ \cos(308.86^o) +i \sin(308.86^o)\right] ~~ \approx ~~ 0.82~~ - ~~1.02i`

just a quick clarification, notice that if we get the inverse tangent of (-1 / -2) the angle we get will be in the range of ±π/2, that's because that is the range inverse tangent is restricted to, however, our terminal point on the complex plane is on the III Quadrant, not the 1st one, so we use the reference angle on the III Quadrant, and that is about 206.57°.