Question

A point $P\left(Z\right)$ moves along the $y$-axis in the argand diagram.

A second point $Q\left(k\right)$ moves along a curve $L$ such that $k=\frac{(2+z)}{(2-z)}$.

$R\left(\lambda \right)$ is a third point such that $\lambda =\sum _{f=0}^{200}{(\cos \frac{3\pi }{2}-i\sin \frac{3\pi }{2})}^f$, then

• A. $\lambda$ lies on $y$-axis
• B. $\lambda$ does not lies on $x$-axis
• C. $\lambda$ lies on curve $L$
• D. None of these

Answer 1

Answer: (C)

$\lambda$ lies on curve $L$

$\lambda =\sum _{f=0}^{200}{(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})}^{f/2}$

$\because \cos \frac{3\pi }{2}=0,\sin \frac{3\pi }{2}=-1$

$\Rightarrow \lambda =\sum _{f=0}^{200}{(-i)}^{f/2}$

$\therefore \lambda =1+\sum _{f=1}^8(-i{)}^{f/2}+\sum _{f=9}^{16}(-i{)}^{f/2}+\cdots +\sum _{f=193}^{200}(-i{)}^{f/2}$

Now, consider $\sum _{f=1}^8(-i{)}^{f/2}$, this term is zero.

Similarly, $\sum _{f=9}^{16}(-i{)}^{f/2}=0$ and so on...

$\therefore \lambda =1$

or

$\lambda =1+0i$

$\Rightarrow \lambda$ lies on $x-$ axis.

Consider point $P$, $z=x+iy$

$\therefore k=\frac{2+iy}{2-iy}$

$|k|=1,\Rightarrow |L|$ is a circle, $|k|=1$

Also $|\lambda |=1$

So, $\lambda$ which also lies on curve $L$