De-moivre’s theorem - Practice Questions & MCQ

De-moivre’s theorem is based on the Euler form representation of complex numbers.  According to De-moivre’s theorem  1.  If $\mathrm{n} \in \mathbb{I}$ (integers), $\theta \in \mathbb{R}$ and $i=\sqrt{-1}$, then $ (\cos \theta+i \sin \theta)^{\mathbf{n}}=\cos \mathbf{n} \theta+i \sin \mathbf{n} \theta $ Proof: We know that, $\mathrm{e}^{i \theta}=\cos \theta+i \sin \theta$ now, $ \begin{aligned} & \Rightarrow\left(\mathrm{e}^{i \theta}\right)^{\mathrm{n}}=(\cos \theta+i \sin \theta)^{\mathrm{n}} \\ & \Rightarrow \mathrm{e}^{i(\mathrm{n} \theta)}=(\cos \theta+i \sin \theta)^{\mathrm{n}} \\ & \Rightarrow \cos \mathrm{n} \theta+i \sin \mathrm{n} \theta=(\cos \theta+i \sin \theta)^n \end{aligned} $ Note:  If n is a rational number which is not an integer and n= p/q , where HCF (p,q) =1 and q>0, then zn will have q values. One of the values will be cos n? + i sin n? 2. If $\theta_1, \theta_2, \theta_3 \ldots \ldots ., \theta_{\mathrm{n}} \in \mathbb{R}$, then $ \begin{gathered} \left(\cos \theta_1+i \sin \theta_1\right)\left(\cos \theta_2+i \sin \theta_2\right)\left(\cos \theta_3+i \sin \theta_3\right) \ldots \ldots \ldots \ldots \ldots \ldots\left(\cos \theta_{\mathrm{n}}+i \sin \theta_{\mathrm{n}}\right) \\ =\cos \left(\theta_1+\theta_2+\theta_3+\ldots \ldots .+\theta_{\mathrm{n}}\right)+i \sin \left(\theta_1+\theta_2+\theta_3+\ldots \ldots .+\theta_{\mathrm{n}}\right) \end{gathered} $ Proof: $\theta_1, \theta_2, \theta_3 \ldots \ldots \ldots . \theta_{\mathrm{n}} \in \mathbb{R}$ and $\mathrm{i}=\sqrt{-1}$, then by Euler's formula $\mathrm{e}^{\mathrm{i} \theta}=\cos \theta+\mathrm{i} \sin \theta$ $\left(\cos \theta_1+\mathrm{i} \sin \theta_1\right)\left(\cos \theta_2+\mathrm{i} \sin \theta_2\right)\left(\cos \theta_3+\mathrm{i} \sin \theta_3\right) \ldots \ldots$. $ \begin{aligned} & \ldots \ldots\left(\cos \theta_{\mathrm{n}}+\mathrm{i} \sin \theta_{\mathrm{n}}\right)=\mathrm{e}^{\mathrm{i} \theta_1} \cdot \mathrm{e}^{\mathrm{i} \theta_2} \cdot \mathrm{e}^{\mathrm{i} \theta_3} \cdot \ldots \ldots \ldots \cdot \mathrm{e}^{\mathrm{i} \theta_{\mathrm{n}}} \\ & =\mathrm{e}^{\mathrm{i}\left(\theta_1+\theta_2+\theta_3+\ldots \ldots+\theta_{\mathrm{n}}\right)} \\ & =\cos \left(\theta_1+\theta_2+\theta_3+\ldots \ldots+\theta_{\mathrm{n}}\right)+\mathrm{i} \sin \left(\theta_1+\theta_2+\theta_3+\ldots \ldots+\theta_{\mathrm{n}}\right) \end{aligned} $