Question

# ^^20(3)(a)^^20\mleft|je^{j\theta}+2\mright|=\sqrt{4+\sin^{2} \theta-2 \sin\theta+\cos^{2} \theta} \left|j e^{j \theta}+2\right|=\sqrt{5-2 \sin \theta} equation i & ii implies - \left|2 e^{j \theta}-j\right|=\left|j e^{j \theta}+2\right| \Rightarrow \quad\left|\frac{2 e^{j \theta}-j}{j e^{j \theta}+2}\right|=1 \Rightarrow \operatorname{Im}\left(z+\frac{\bar{z}}{z \bar{z}}\right)=0 \Rightarrow\operatorname{Im}(z)+\operatorname{Im}\mleft(z+\frac{\bar{z}}{|z|^2}\mright)=0 \begin{array}{rl}a_1 & z\bar{z}=|z|^2 \\ & \text{^^20also^^20},\operatorname{Im}(z)=-\operatorname{Jm}(z)\end{array} \Rightarrow \operatorname{Im}(z)-\frac{\operatorname{Im}(z)}{|z|^{2}}=0 \Rightarrow \quad \operatorname{Im}(z) \cdot\left[1-\frac{1}{|z|^{2}}\right]=0 \Rightarrow \text { either } \operatorname{Im}(z)=0 \text { or } 1-\frac{1}{|z|^{2}}=0 either z is real or IzI^2 =1 i.e., IzI=1 3) c) as we know that - (a \cdot b)^{m}=a^{m} \cdot b^{m} \quad \text { iff atleat one } a>0 \text { or } b>0 but here (-1)^{1 / 2}(-1)^{1 / 2}=(-1 \times-1)^{1 / 2} This step is false hence given statement is wrong. \Rightarrow \quad \operatorname{Im}\left(z+\frac{1}{z}\right)=0 3) b) z+1/z is real.  Fig: 1  Fig: 2  Fig: 3  Fig: 4  Fig: 5  Fig: 6  Fig: 7  Fig: 8  Fig: 9  Fig: 10  Fig: 11  Fig: 12  Fig: 13  Fig: 14  Fig: 15  Fig: 16  Fig: 17  Fig: 18  Fig: 19  Fig: 20  Fig: 21