SATHEE: Complex Numbers 2 Question 8

Complex Numbers 2 Question 8

8. Let complex numbers $α$ and $1 / \bar{α}$ lies on circles ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ , respectively.

If $z_{0} = x_{0} + i y_{0}$ satisfies the equation $2 {| z_{0} |}^{2} = r^{2} + 2$ , then $| α |$ is equal to

(2013 Adv.)

(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{2}$

(d) $\frac{1}{3}$

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Answer:

Correct Answer: 8. (c)

Solution:

PLAN Intersection of circles, the basic concept is to solve the equations simultaneously and using properties of modulus of complex numbers.

Formula used $| z |^{2} = z \cdot \bar{z}$

and

$\begin{aligned} {| z_{1} - z_{2} |}^{2} & = (z_{1} - z_{2}) ({\bar{z}}_{1} - {\bar{z}}_{2}) \\ = {| z_{1} |}^{2} - z_{1} {\bar{z}}_{2} - z_{2} {\bar{z}}_{1} + {| z_{2} |}^{2} \end{aligned}$

Here, ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$

and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ can be written as,

${| z - z_{0} |}^{2} = r^{2} and {| z - z_{0} |}^{2} = 4 r^{2}$

Since, $α$ and $\frac{1}{\bar{α}}$ lies on first and second respectively.

$∴ {| α - z_{0} |}^{2} = r^{2}$ and ${| \frac{1}{\bar{α}} - z_{0} |}^{2} = 4 r^{2}$

$\Rightarrow (α - z_{0}) (\bar{α} - {\bar{z}}_{0}) = r^{2}$

$\Rightarrow | α |^{2} - z_{0} \bar{α} - {\bar{z}}_{0} α + {| z_{0} |}^{2} = r^{2}$

and ${| \frac{1}{\bar{α}} - z_{0} |}^{2} = 4 r^{2}$

$\Rightarrow \frac{1}{\bar{α}} - z_{0} \frac{1}{α} - {\bar{z}}_{0} = 4 r^{2}$

$\Rightarrow \frac{1}{| α |^{2}} - \frac{z_{0}}{α} - \frac{{\bar{z}}_{0}}{\bar{α}} + {| z_{0} |}^{2} = 4 r^{2}$

Since, $| α |^{2} = α \cdot α$

$\Rightarrow \frac{1}{| α |^{2}} - \frac{z_{0} \cdot \bar{α}}{| α |^{2}} - \frac{{\bar{z}}_{0}}{| α |^{2}} \cdot α + {| z_{0} |}^{2} = 4 r^{2}$

$\Rightarrow 1 - z_{0} \bar{α} - {\bar{z}}_{0} α + | α |^{2} {| z_{0} |}^{2} = 4 r^{2} | α |^{2}$

On subtracting Eqs. (i) and (ii), we get

$\begin{aligned} (| α |^{2} - 1) + {| z_{0} |}^{2} (1 - | α |^{2}) & = r^{2} (1 - 4 | α |^{2}) \\ \Rightarrow (| α |^{2} - 1) (1 - {| z_{0} |}^{2}) & = r^{2} (1 - 4 | α |^{2}) \\ \Rightarrow (| α |^{2} - 1) 1 - \frac{r^{2} + 2}{2} & = r^{2} (1 - 4 | α |^{2}) \\ Given, & {| z_{0} |}^{2} & = \frac{r^{2} + 2}{2} \end{aligned}$

$\begin{aligned} \Rightarrow (| α |^{2} - 1) \cdot \frac{r^{2}}{2} = r^{2} (1 - 4 | α |^{2}) \\ \Rightarrow | α |^{2} - 1 = - 2 + 8 | α |^{2} \\ \Rightarrow 7 | α |^{2} = 1 \\ ∴ | α | = 1 / \sqrt{7} \end{aligned}$

Complex Numbers 2 Question 8

8. Let complex numbers $α$ and $1 / \bar{α}$ lies on circles ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ , respectively.

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Complex Numbers 2 Question 8

8. Let complex numbers α and 1/α¯ lies on circles (x−x0)2+(y−y0)2=r2 and (x−x0)2+(y−y0)2=4r2, respectively.

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8. Let complex numbers $α$ and $1 / \bar{α}$ lies on circles ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ , respectively.