Question Number 94080 by Rio Michael last updated on 16/May/20
$$\mathrm{3}.\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:{z}\:\mathrm{which}\:\mathrm{satisfy}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:{z}^{\mathrm{3}} \:=\:\mathrm{8}{i}\:,\:\mathrm{giving}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{form}\:{a}\:+\:{bi}\:\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{real}. \\ $$
Answered by mr W last updated on 17/May/20
![say z=re^(iθ) =r (cos θ+i sin θ) r^3 e^(3θi) =8i=2^3 e^((π/2)i) ⇒r=2 ⇒3θ=(π/2)+2kπ ⇒θ=(π/6)+((2kπ)/3) with k=0,1,2 ⇒z=2e^(((π/6)+((2kπ)/3))i) =2[cos ((π/6)+((2kπ)/3))+i sin ((π/6)+((2kπ)/3))] =(√3)+i, −(√3)+i, −2i](https://www.tinkutara.com/question/Q94159.png)
$${say}\:{z}={re}^{{i}\theta} ={r}\:\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${r}^{\mathrm{3}} {e}^{\mathrm{3}\theta{i}} =\mathrm{8}{i}=\mathrm{2}^{\mathrm{3}} {e}^{\frac{\pi}{\mathrm{2}}{i}} \\ $$$$\Rightarrow{r}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:\Rightarrow\theta=\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\:{with}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\Rightarrow{z}=\mathrm{2}{e}^{\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right){i}} =\mathrm{2}\left[\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)+{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\right] \\ $$$$=\sqrt{\mathrm{3}}+{i},\:−\sqrt{\mathrm{3}}+{i},\:−\mathrm{2}{i} \\ $$
Commented by Rio Michael last updated on 17/May/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$