Question

Write the complex number $z= \frac{(1-\sqrt{3}i)^{2015} }{[\sqrt{2}cis(\frac{7\pi}{6})]^{4028}}$ in an exponential form.

Answer 1

Step-by-step explanation:

We have,

$z= \frac{(1-\sqrt{3}i)^{2015} }{[\sqrt{2}cis(\frac{7\pi}{6})]^{4028}}$

$\implies \: z= \frac{ ({2})^{2015} \dfrac{(1-\sqrt{3}i)^{2015}}{ {2}^{(2015)} } }{[\sqrt{2} {e}^{\frac{7\pi}{6}}]^{4028}}$

$\implies \: z= \frac{ ({2})^{2015} \bigg( \dfrac{1}{2} - \dfrac{ \sqrt{3} }{2} i \bigg)^{2015} }{(2)^{2014} ({e}^{\frac{7\pi}{6}}) ^{4028}}$

$\implies \: z= \frac{ ({2})^{2015 - 2014} \bigg \{ \cos \bigg( \dfrac{11\pi}{6} \bigg) + i \sin \bigg(\dfrac{ 11\pi }{6} \bigg) \bigg \}^{2015} }{ {e}^{(\frac{7\pi}{6} \times 4028)}}$

$\implies \: z= \frac{2 \{ {e}^{ \frac{11\pi}{6} } \}^{2015} }{ {e}^{(\frac{7\pi}{6} \times 4028)}}$

$\implies \: z= \frac{2 {e}^{ \frac{11\pi}{6} \times 2015 }}{ {e}^{(\frac{7\pi}{6} \times 4028)}}$

$\implies \: z= 2 {e}^{ \frac{11\pi}{6} \times 2015 - \frac{7\pi}{6} \times 4028}$

$\implies \: z= 2 {e}^{ \frac{\pi}{6}(11 \times 2015 - 7 \times 4028)}$

$\implies \: z= 2 {e}^{ \frac{\pi}{6}(22165 - 28196)}$

$\implies \: z= 2 {e}^{ \frac{\pi}{6}( - 6031)}$

$\implies \: z= 2 {e}^{ - \frac{\pi}{6} \times 6031}$

$\implies \: z= 2 {e}^{ - (1005)\pi - \frac{\pi}{6} }$

$\implies \: z= 2 \bigg[ \cos \bigg \{ - (1005)\pi - \frac{\pi}{6} \bigg \} + i \sin \bigg \{ - (1005)\pi - \frac{\pi}{6} \bigg \} \bigg ]$

$\implies \: z= 2 \bigg[ \cos \bigg \{ (1005)\pi + \frac{\pi}{6} \bigg \} - i \sin \bigg \{ (1005)\pi + \frac{\pi}{6} \bigg \} \bigg ]$

$\implies \: z= 2 \bigg[ - \cos \bigg (\frac{\pi}{6} \bigg) + i \sin \bigg ( \frac{\pi}{6} \bigg ) \bigg ]$

$\implies \: z= 2 \bigg[ - \frac{ \sqrt{3} }{2} + i . \frac{1}{2} \bigg ]$

$\implies \: z= 2 \bigg[ - \frac{ \sqrt{3} }{2} + \frac{i}{2} \bigg ]$

$\implies \: z= 2 \bigg[ \frac{ - \sqrt{3} + i}{2} \bigg ]$

$\implies \: z= - \sqrt{3} + i$