SATHEE: Complex Numbers 3 Question 12

Complex Numbers 3 Question 12

12. Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $(z - z_{1}) / (z - z_{2})$ is $π / 4$ , then prove that $| z - 7 - 9 i | = 3 \sqrt{2}$ .

(1991, 4M)

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Solution:

Since, $z_{1} = 10 + 6 i, z_{2} = 4 + 6 i$

and arg $\frac{z - z_{1}}{z - z_{2}} = \frac{π}{4}$ represents locus of $z$ is a circle shown as from the figure whose centre is $(7, y)$ and $∠ A O B = 90^{\circ}$ , clearly $O C = 9 \Rightarrow O D = 6 + 3 = 9$

$∴$ Centre $= (7, 9)$ and radius $= \frac{6}{\sqrt{2}} = 3 \sqrt{2}$

$\Rightarrow$ Equation of circle is $| z - (7 + 9 i) | = 3 \sqrt{2}$

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Complex Numbers 3 Question 12

12. Let z1=10+6i and z2=4+6i. If z is any complex number such that the argument of (z−z1)/(z−z2) is π/4, then prove that |z−7−9i|=32.

Solution:

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12. Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $(z - z_{1}) / (z - z_{2})$ is $π / 4$ , then prove that $| z - 7 - 9 i | = 3 \sqrt{2}$ .