Mathematics - Complex Number Question with Solution | TestHub

Clearly ${\alpha }_1=1$

$\begin{array}{rlrlr}& {\alpha }_2=\alpha & {}^{}& & \end{array}$

${\alpha }_3=\alpha 2$

${\alpha }_4={\alpha }^3$

${\alpha }_5={\alpha }^4$

where $\alpha ={\mathrm{e}}^{\mathrm{i}2\pi /5}$

$\begin{array}{rlrlrl}& \therefore \lambda ={\alpha }_3^{1001}={\left({\alpha }^2\right)}^{1001}={\alpha }^{2002}={\alpha }^{5\times 400+2}={\alpha }^2& & & & \end{array}$

$\mu ={\left({\alpha }_4\right)}^{669+1/3}={\left({\alpha }^3\right)}^{(669+1/3)}={\alpha }^{2008}={\alpha }^3$

$v={\left({\alpha }_5\right)}^{503+1/2}={\left({\alpha }^4\right)}^{503+1/2}$

$={\alpha }^{2014}\cdot ={\alpha }^{5.402+4}={\alpha }^4$

Also sum of $2011{}^{\text{th }}$ power of roots of unity is 0 So, $1+{\alpha }^{2011}+{\lambda }^{2011}+{\mu }^{2011}+{\mathrm{v}}^{2011}=0$

${\lambda }^{2011}+{\mu }^{2011}+v^{2011}=-\left(1+{\alpha }^{2011}\right)$

${\lambda }^{2011}+{\mu }^{2011}+v^{2011}=-(1+\alpha )$

$\left|{\lambda }^{2011}+{\mu }^{2011}+v^{2011}\right|=\left|-\left(1+e^{i2\pi /5}\right)\right|$

$=|1+\cos 2\pi /5+i\sin 2\pi /5|=\mid 2\cos \pi /5(\cos \pi /5+i\sin \pi /5)$

$=2\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{2}$

$\begin{array}{rlrlrlr}& & & & & & \left[\left|{\lambda }^{2011}+{\mu }^{2011}+{\mathrm{v}}^{2011}\right|\right]=1\end{array}$