Imaginary part of ({3}-i)²016}+(-{3}-i)²019} is

Correct Answer is : - 220192²019}22019

Correct Answer is : - 220192²019}22019

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TS EAMCET 03 May 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

Imaginary part of

(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}

Solution:

(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}

=2^{2016}\left(\cos \;\frac{\pi}{6}-i \sin \;\frac{\pi}{6}\right)^{2016}-2^{2019}\left(\cos \;\frac{\pi}{6}+i \sin \;\frac{\pi}{6}\right)^{2019}

=\begin{bmatrix} 2^{2016}\left(\cos \;\frac{2016 \pi}{6}-i \sin \;\frac{2016 \pi}{6}\right) \\ -2^{2019}\left(\cos \;\frac{2019 \pi}{6}+i \sin \;\frac{2019 \pi}{6}\right) \end{bmatrix}

= \;\; 2^{2016}(\cos 336\pi - i \sin 336\pi)-

\& 2^{2019}\left[\cos \left(336\pi + \;\frac{\pi}{2}\right)+i \sin \left(336\pi + \;\frac{\pi}{2}\right)\right]

= \;\; 2^{2016}-i 2^{2019}

Therefore, the imaginary part is

-2^{2019}

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