Mathematics - Complex Number Question with Solution | TestHub

Let $z=e^{i\theta }\cdot 2\cos \theta =\left(z+\frac{1}{z}\right)$ and $2i\sin \theta =\left(z-\frac{1}{z}\right)$

Now : $(2\cos \theta {)}^3(2i\sin \theta {)}^5={\left(z+\frac{1}{z}\right)}^3{\left(z-\frac{1}{z}\right)}^5$ $=\left(z^8-\frac{1}{z^8}\right)-2\left(z^6-\frac{1}{z^6}\right)-2\left(z^4-\frac{1}{z^4}\right)+6\left(z^2-\frac{1}{z^2}\right)$

Compare $\mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=-2,\mathrm{d}=6$

$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=1+2-2+6=7$