Menu Close

if-3-2-1-2-i-find-z-11-




Question Number 146467 by mathdanisur last updated on 13/Jul/21
if    ((√3)/2) + (1/2) i    find   z^(11)  = ?
$${if}\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{{i}}\:\:\:\:{find}\:\:\:\boldsymbol{{z}}^{\mathrm{11}} \:=\:? \\ $$
Answered by gsk2684 last updated on 13/Jul/21
z^(11)  = (cis (Π/6))^(11) = cis(((11Π)/6))  =cis(2Π−(Π/6))  =cis(−(Π/6))  =((√3)/2)−(i/2)
$${z}^{\mathrm{11}} \:=\:\left({cis}\:\frac{\Pi}{\mathrm{6}}\right)^{\mathrm{11}} =\:{cis}\left(\frac{\mathrm{11}\Pi}{\mathrm{6}}\right) \\ $$$$={cis}\left(\mathrm{2}\Pi−\frac{\Pi}{\mathrm{6}}\right) \\ $$$$={cis}\left(−\frac{\Pi}{\mathrm{6}}\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thanks Ser, but cis.? cos.?
$${thanks}\:{Ser},\:{but}\:{cis}.?\:{cos}.? \\ $$
Commented by gsk2684 last updated on 13/Jul/21
cis θ = cos θ + i sin θ
$${cis}\:\theta\:=\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\:\theta \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thanks Ser
$${thanks}\:{Ser} \\ $$
Answered by puissant last updated on 13/Jul/21
Z=cos(π/6)+i sin(π/6)=e^(i(π/6))   Z^(11) =(e^(i(π/6)) )^(11) =e^(i((11π)/6)) = cos((11π)/6)+i sin((11π)/6)  ⇒ Z=((√3)/2)−(1/2)i
$$\mathrm{Z}=\mathrm{cos}\frac{\pi}{\mathrm{6}}+\mathrm{i}\:\mathrm{sin}\frac{\pi}{\mathrm{6}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\mathrm{Z}^{\mathrm{11}} =\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} \right)^{\mathrm{11}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{11}\pi}{\mathrm{6}}} =\:\mathrm{cos}\frac{\mathrm{11}\pi}{\mathrm{6}}+\mathrm{i}\:\mathrm{sin}\frac{\mathrm{11}\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{Z}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thank you Ser  cos((11π)/6)+isin((11π)/6)⇒z^(11) =((√3)/2)−(1/2)i.?
$${thank}\:{you}\:{Ser} \\ $$$${cos}\frac{\mathrm{11}\pi}{\mathrm{6}}+{isin}\frac{\mathrm{11}\pi}{\mathrm{6}}\Rightarrow{z}^{\mathrm{11}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i}.? \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thanks Ser
$${thanks}\:{Ser} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *