Complex Numbers Ques 36
Match the statements of Column I with those of Column II.
Here, $z$ takes values in the complex plane and $\operatorname{Im}(z)$ and $\operatorname{Re}(z)$ denote respectively, the imaginary part and
| Column I | Column II | |
|---|---|---|
| A. | The set of points $z$ satisfying $|z-i| z||=|z+i| z||$ is contained in or equal to |
p. an ellipse with eccentricity 4/5 |
| B. | The set of points $z$ satisfying $|z+4|+|z-4|=0$ is contained in or equal to |
q. the set of points $z$ satisfying $\operatorname{Im}(z)=0$ |
| C. | If $|w|=2$, then the set of points $z=w-\frac{1}{w}$ is contained in or equal to |
r. the set of points $z$ satisfying $|\operatorname{lm}(z)| \leq 1$ |
| D. | If $|w|=1$, then the set of points $z=w+\frac{1}{w}$ is contained in or equal to |
s. the set of points t. satisfying $|\operatorname{Re}(z)| \leq 2$ the set of points $z$ satisfying $|z| \leq 3$ |
(2010)
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Answer:
Correct Answer: 36.$A → q, r ; B →p; C →p, s, t ; D →q, r, s, t$
Solution:
Formula:
- A. Let $\quad z=x+i y$
$\Rightarrow$ we get $\quad y \sqrt{x^{2}+y^{2}}=0$
$\Rightarrow$ $y=0$
$\Rightarrow$ $I _m(z)=0$
B. We have
$\Rightarrow \quad 2 a e =8,2 a=10 $
$\Rightarrow \quad 10 e =8 $
$\Rightarrow \quad e =\frac{4}{5} $
$\Rightarrow \quad b^{2} =25 \quad (1-\frac{16}{25})=9 $
$\therefore \quad \frac{x^{2}}{25}+\frac{y^{2}}{9} =1$
C. Let $w=2(\cos \theta+i \sin \theta)$
$\therefore \quad z=2(\cos \theta+i \sin \theta)-\frac{1}{2(\cos \theta+i \sin \theta)}$
$ \begin{aligned} & =2(\cos \theta+i \sin \theta)-\frac{1}{2}(\cos \theta-i \sin \theta) \\ & =\frac{3}{2} \cos \theta+\frac{5}{2} i \sin \theta \end{aligned} $
Let $\quad z=x+i y$
$\Rightarrow \quad x=\frac{3}{2} \cos \theta$ and $y=\frac{5}{2} \sin \theta$
$\Rightarrow \quad (\frac{2 x}{3})^2+(\frac{2 y}{5})^{2}=1$
$\Rightarrow \quad \frac{x^{2}}{9 / 4}+\frac{y^{2}}{25 / 4}=1$
$\therefore \quad e=\sqrt{1-\frac{9 / 4}{25 / 4}}=\frac{4}{5}$
D. Let $\quad w=\cos \theta+i \sin \theta$
Then, $\quad z=x+i y=\cos \theta+i \sin \theta+\frac{1}{\cos \theta+i \sin \theta}$
$ =2 \cos \theta $
$\Rightarrow \quad x=2 \cos \theta, y=0$
BETA
