Question

Find the real and imaginary parts of (-1-i)⁷+(-1+i)⁷

Answer 1

Answer:

Step-by-step explanation:

We have,

$z=\left(-1-i\right)^{7}+\left(-1+i\right)^{7}$

$\implies\,z=\left(\sqrt{2}\right)^{7}\cdot\left(-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i\right)^{7}+\left(\sqrt{2}\right)^{7}\cdot\left(-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)^{7}$

$\implies\,z=\left(\sqrt{2}\right)^{7}\left\{\left(-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i\right)^{7}+\left(-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)^{7}\right\}$

$\implies\,z=\left(\sqrt{2}\right)^{7}\left[\left\{\cos\left(\dfrac{5\pi}{4}\right)+i\sin\left(\dfrac{5\pi}{4}\right)\right\}^{7}+\left\{\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)\right\}^{7}\right]$

Using De moivre's theorem,

$\implies\,z=\left(\sqrt{2}\right)^{7}\left[\cos\left(\dfrac{35\pi}{4}\right)+i\sin\left(\dfrac{35\pi}{4}\right)+\cos\left(\dfrac{21\pi}{4}\right)+i\sin\left(\dfrac{21\pi}{4}\right)\right]$

$\implies\,z=\left(\sqrt{2}\right)^{7}\left[\cos\left(9\pi-\dfrac{\pi}{4}\right)+i\sin\left(9\pi-\dfrac{\pi}{4}\right)+\cos\left(5\pi+\dfrac{\pi}{4}\right)+i\sin\left(5\pi+\dfrac{\pi}{4}\right)\right]$

$\implies\,z=\left(\sqrt{2}\right)^{7}\left[-\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)-\cos\left(\dfrac{\pi}{4}\right)-i\sin\left(\dfrac{\pi}{4}\right)\right]$

$\implies\,z=\left(\sqrt{2}\right)^{7}\left[-2\cos\left(\dfrac{\pi}{4}\right)\right]$

$\implies\,z=-2\cdot\left(\sqrt{2}\right)^{7}\cdot\dfrac{1}{\sqrt{2}}$

$\implies\,z=-2\cdot\left(\sqrt{2}\right)^{6}$

$\implies\,z=-2\cdot\left(2\right)^{3}$

$\implies\,z=-16$