If z = 3 + i 2 then z 101 + i 103 105 =

Given,

$z=\frac{\sqrt{3}+i}{2}$

$=\left(\frac{\sqrt{3}}{2}\right)+i\left(\frac{1}{2}\right)$

$=\cos \left(\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\frac{\mathrm{\pi }}{6}\right)$

Then,

$z^{101}={\left[\cos \left(\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\frac{\mathrm{\pi }}{6}\right)\right]}^{101}$

$=\cos \left(\frac{101\mathrm{\pi }}{6}\right)+i\sin \left(\frac{101\mathrm{\pi }}{6}\right)$

$=\cos \left(\frac{101}{6}\mathrm{\pi }\right)+i\sin \left(\frac{101}{6}\mathrm{\pi }\right)$

$=\left\{\cos \left(17\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)+i\sin \left(17\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\right\}$

$=\left\{-\cos \left(\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\frac{\mathrm{\pi }}{6}\right)\right\}$ 

$=\frac{-\sqrt{3}+i}{2}$

And, $i^{103}=i^{4\times 25+3}=i^3=-i$

Therefore,

${\left(z^{101}+i^{103}\right)}^{105}={\left(\frac{-\sqrt{3}}{2}+\frac{i}{2}-i\right)}^{105}$

$={\left(\frac{-\sqrt{3}}{2}-\frac{i}{2}\right)}^{105}$

$=-{\left(\frac{\sqrt{3}+i}{2}\right)}^{105}$

$=-z^{105}$

Now,

$-z^{105}=-{\left(\frac{\sqrt{3}+i}{2}\right)}^{105}$

                $=-{\left\{\cos \left(\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\frac{\mathrm{\pi }}{6}\right)\right\}}^{105}$

$=-\left\{\cos \left(\frac{105\mathrm{\pi }}{6}\right)+i\sin \left(\frac{105\mathrm{\pi }}{6}\right)\right\}$

$=-\left\{\cos \left(18\mathrm{\pi }-\frac{3\mathrm{\pi }}{6}\right)+i\sin \left(18\mathrm{\pi }-\frac{3\mathrm{\pi }}{6}\right)\right\}$

 $=-\left\{\cos \left(\frac{3\mathrm{\pi }}{6}\right)-i\sin \left(\frac{3\mathrm{\pi }}{6}\right)\right\}$

 $=-\left\{\cos \left(\frac{\mathrm{\pi }}{2}\right)-i\sin \left(\frac{\mathrm{\pi }}{2}\right)\right\}$

 $=-\left(-i\right)$

$=i$