If $$n=0$$  we get the answer as 2
If $$n$$ is even and $$n$$ is simply a multiple of 2 but not a multiple of 4, we get the answer as $$-$$ 2
If $$n$$ is odd,
Let $$n=3$$  we get
$$\eqalign{ & - i - \frac{1}{i} \cr & = \frac{{1 - 1}}{i} \cr & = 0 \cr & n = 5{\text{ we get}} \cr & i + \frac{1}{i} \cr & = 0 \cr} $$
Hence in total there are only 3 values, $$ - 2,\,0,\,2$$

$$\eqalign{ & {\left| {\alpha + \beta } \right|^2} = \left( {\alpha + \beta } \right)\left( {\overline {\alpha + \beta } } \right) = \left( {\alpha + \beta } \right)\left( {\overline \alpha + \overline \beta } \right) = \alpha \overline \alpha + \beta \overline \beta + \alpha \overline \beta + \overline \alpha \beta \cr & {\left| {\alpha + \beta } \right|^2} = {\left| \alpha \right|^2} + {\left| \beta \right|^2} + \alpha \overline \beta + \overline \alpha \beta \cr & {\left| {\alpha - \beta } \right|^2} = \left( {\alpha - \beta } \right)\left( {\overline \alpha - \overline \beta } \right) = \alpha \overline \alpha + \beta \overline \beta - \alpha \overline \beta - \overline \alpha \beta \cr & {\left| {\alpha - \beta } \right|^2} = {\left| \alpha \right|^2} + {\left| \beta \right|^2} - \alpha \overline \beta - \overline \alpha \beta . \cr} $$

$$\eqalign{ & {\text{If }}\left| {\frac{{z - 1}}{{z + 1}}} \right| = r\,{\text{then }}\frac{{z - 1}}{{z + 1}} = r\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right) = r\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right) \cr & {\text{or, }}\frac{{\left( {x - 1} \right) + iy}}{{\left( {x + 1} \right) + iy}} = \frac{r}{2} + i\frac{{r\sqrt 3 }}{2} \cr & {\text{or, }}\left( {x - 1} \right) + iy = \frac{r}{2}\left( {x + 1} \right) - \frac{{yr\sqrt 3 }}{2} + i\left\{ {\frac{{ry}}{2} + \frac{{r\sqrt 3 }}{2}\left( {x + 1} \right)} \right\} \cr} $$
\[\left. \begin{array}{l} \Rightarrow \,\,x - 1 = \frac{r}{2}\left( {x + 1} \right) - \frac{{yr\sqrt 3 }}{2}\\ \,\,\,\,\,\,\,y = \frac{{ry}}{2} + \frac{{r\sqrt 3 }}{2}\left( {x + 1} \right) \end{array} \right\}\]
$$ \Rightarrow \,\,\frac{{x - 1}}{y} = \frac{{x + 1 - y\sqrt 3 }}{{y + \sqrt 3 \left( {x + 1} \right)}}.$$
On simplification, $$\sqrt 3 \left( {{x^2} + {y^2}} \right) - 2y - \sqrt 3 = 0,$$      which is a circle.

$$\eqalign{ & {1^{\frac{1}{4}}} = {\left( {\cos 2\pi r + i\sin 2\pi r} \right)^{\frac{1}{4}}} = \cos \frac{{\pi r}}{2} + i\sin \frac{{\pi r}}{2};r = 0,1,2,3 \cr & \therefore \,\,{1^{\frac{1}{4}}} = 1,i, - 1, - i. \cr & \therefore \,\,{\text{value}} = {1^2} + {i^2} + {\left( { - 1} \right)^2} + {\left( { - i} \right)^2}. \cr} $$

The given eq. implies that the difference between the distances of the moving point from two fixed points $$\left( {1 - i} \right)$$  and $$\left( {2 + i} \right)$$  is constant using the property of the hyperbola that the difference between the focal distances of any point on the curve is constant, the locus in reference is therefore a hyperbola.

$$\eqalign{ & {z^3} + \frac{{3{{\left| z \right|}^2}}}{z} = 0, \cr & \Rightarrow \,{z^3} + \frac{{3z.\bar z}}{z} = 0 \cr & \Rightarrow \,{z^3} + 3\bar z = 0 \cr & {\text{Let, }}\,z = r{e^{i\theta }} \cr & \Rightarrow \,{r^3}{e^{i3\theta }} + 3r{e^{ - i\theta }} = 0 \cr & \Rightarrow \,{e^{i4\theta }} = - 1\,\,\,\left[ {\because \,r = \sqrt 3 } \right] \cr & \Rightarrow \,\cos \,4\theta + i\,\sin \,4\theta = - 1 \cr & \Rightarrow \,\cos \,4\theta = - 1\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{Now, }}\,0 \leqslant \theta < 2\pi \cr & \Rightarrow \,0 \leqslant 4\theta < 8\pi \cr & \therefore \,\theta = \pi ,3\pi ,5\pi ,7\pi \cr} $$

$$\eqalign{ & {\text{Given }}\left| z \right| = 1,arg\,z = \theta \cr & {\text{As we know, }}\overline z = \frac{1}{z} \cr & \therefore \,\,\arg \left( {\frac{{1 + z}}{{1 + \overline z }}} \right) \cr & = \arg \left( {\frac{{1 + z}}{{1 + \frac{1}{z}}}} \right) \cr & = \arg \left( z \right) = \theta. \cr} $$

$$\eqalign{ & {\left( {1 + \omega } \right)^7} = A + B\omega ;\,\,\,\,\,\,\,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega \cr & - {\omega ^2} = A + B\omega ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + \omega = A + B\omega \cr & \Rightarrow \,\,A = 1,B = 1. \cr} $$

$$\eqalign{ & z = 1 + i\,\tan \,\alpha = r\left( {\cos \theta + i\,\sin \theta } \right) \cr & \Rightarrow \,r\,\cos \,\theta = 1,r\,\sin \,\theta = \tan \,\alpha \cr & \Rightarrow \,{r^2} = {\sec ^2}\alpha \cr & \Rightarrow \,r = \left| {\sec \,\alpha } \right| = \frac{1}{{\left| {\cos \,\alpha } \right|}} \cr & {\text{Since}}, - \pi < \alpha < - \frac{\pi }{2} \cr & \Rightarrow \,\cos \,\alpha \, < 0 \cr & \Rightarrow \,\left| {\cos \,\alpha } \right| = - \cos \,\alpha \cr & \therefore \,r = \frac{1}{{ - \cos \,\alpha }}. \cr & {\text{Further,}}\,\,{\text{we}}\,{\text{get,}} \cr & \cos \,\theta = - \cos \,\alpha = \cos \left( {\pi + \alpha } \right) \cr & {\text{Now}},\, - \pi < \alpha < - \frac{\pi }{2} \cr & \Rightarrow \,\pi - \pi < \pi + \alpha < \pi - \frac{\pi }{2} \cr & \Rightarrow \,0 < \pi + \alpha < \frac{\pi }{2}\,\,\,\,\,\left[ {{\text{Converted to principal value}}} \right] \cr & \therefore \,\cos \,\theta = \cos \left( {\pi + \alpha } \right) \cr & \Rightarrow \,\theta = \pi + \alpha \cr & {\text{Hence}},\,\,z = \frac{1}{{ - \cos \,\alpha }}\left[ {\cos \left( {\pi + \alpha } \right) + i\,\sin \left( {\pi + \alpha } \right)} \right] \cr} $$

$$\eqalign{ & {\left( {\sin \,x + i\,\cos \,x} \right)^3} \cr & = \,{\sin ^3}x + {\left( i \right)^3}\,{\cos ^3}x + 3i\,\left( {\sin \,x} \right)\,\left( {\cos \,x} \right)\left( {\sin \,x + i\,\cos \,x} \right) \cr & = \,{\sin ^3}x - i\,{\cos ^3}x + 3i\,{\sin ^2}x\,\cos \,x - 3\,\sin \,x\,{\cos ^2}x \cr & = \,\sin \,x\left( {{{\sin }^2}x - 3\,{{\cos }^2}x} \right) + i\,\cos \,x \cr & {\text{Real}}\,{\text{part}}\,{\text{of}}\,{\left( {\sin \,x + i\,\cos \,x} \right)^3} \cr & = \,\sin \;x\left( {{{\sin }^2}x - 3\,{{\cos }^2}x} \right) \cr & = \,\sin \,x\left[ {{{\sin }^2}x - 3\left( {1 - {{\sin }^2}x} \right)} \right] \cr & = \,\sin \,x\left[ {4{{\sin }^2}x - 3} \right] \cr & = \,4{\sin ^3}x - 3\sin \,x \cr & = \, - \left( {3\,\sin \,x - 4\,{{\sin }^3}x} \right) \cr & = \, - \sin \,3x \cr} $$