Question Number 54991 by Tawa1 last updated on 15/Feb/19
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}:\:\:\:\:\:\:\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{16i}\:\:=\:\:\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 15/Feb/19
![z^4 +16i =0 ⇔ z^4 =−16i let z =r e^(iθ) ⇒r^4 e^(i4θ) =16(−i)=16 e^(−((iπ)/2)) ⇒ r^4 =16 and 4θ =−(π/2) +2kπ ⇒r =2 and θ_k =−(π/8) +((kπ)/2) and k ∈[[0,3]] so the roots of this equation are z_k =2 e^(i(−(π/8)+((kπ)/2))) ⇒z_0 =2 e^(−((iπ)/8)) z_1 = 2 e^(i(((3π)/8))) , z_2 =2 e^(i( ((7π)/8))) , z_3 =2 e^(i(((11π)/8))) also we have z^4 +16i =(z−z_0 )(z−z_1 )(z−z_2 )(z−z_3 ).](Q55001.png)
$${z}^{\mathrm{4}} \:+\mathrm{16}{i}\:=\mathrm{0}\:\Leftrightarrow\:{z}^{\mathrm{4}} =−\mathrm{16}{i}\:\:\:\:\:{let}\:{z}\:={r}\:{e}^{{i}\theta} \:\Rightarrow{r}^{\mathrm{4}} \:{e}^{{i}\mathrm{4}\theta} =\mathrm{16}\left(−{i}\right)=\mathrm{16}\:{e}^{−\frac{{i}\pi}{\mathrm{2}}} \:\Rightarrow \\ $$$${r}^{\mathrm{4}} =\mathrm{16}\:{and}\:\mathrm{4}\theta\:=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\Rightarrow{r}\:=\mathrm{2}\:{and}\:\theta_{{k}} =−\frac{\pi}{\mathrm{8}}\:+\frac{{k}\pi}{\mathrm{2}}\:\:{and}\:{k}\:\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]\:{so}\:{the}\:{roots} \\ $$$${of}\:{this}\:{equation}\:{are}\:{z}_{{k}} =\mathrm{2}\:{e}^{{i}\left(−\frac{\pi}{\mathrm{8}}+\frac{{k}\pi}{\mathrm{2}}\right)} \:\Rightarrow{z}_{\mathrm{0}} =\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \\ $$$${z}_{\mathrm{1}} =\:\mathrm{2}\:{e}^{{i}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)} \:\:\:\:,\:\:{z}_{\mathrm{2}} =\mathrm{2}\:{e}^{{i}\left(\:\frac{\mathrm{7}\pi}{\mathrm{8}}\right)} \:\:\:,\:{z}_{\mathrm{3}} =\mathrm{2}\:{e}^{{i}\left(\frac{\mathrm{11}\pi}{\mathrm{8}}\right)} \:{also}\:{we}\:{have} \\ $$$${z}^{\mathrm{4}} \:+\mathrm{16}{i}\:=\left({z}−{z}_{\mathrm{0}} \right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{3}} \right). \\ $$
Commented by Tawa1 last updated on 16/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$
Commented by maxmathsup by imad last updated on 16/Feb/19
$${you}\:{are}\:{welcome}\:{sir}\:. \\ $$