If z₁ is a point on zbar{z} = 1 and z₂ is another point on (4 -3i)z + (4 + 3i)z -15 = 0 , then |

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4-1.Complex numbers

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If $z_1$ is a point on $z\bar{z} = 1$ and $z_2$ is another point on $(4 -3i)z + (4 + 3i)z -15 = 0$, then $|z_1 -z_2|_{min}$ is (where $ i = \sqrt { - 1}$ )

A

$\frac{1}{2}$

B

$2$

C

$\frac{3}{2}$

D

$4$

Solution

$z \bar{z}=1 \Rightarrow x^{2}+y^{2}=1$

$(4-3 i) z+(4+3 i) \bar{z}-15=0$

$\Rightarrow 8 x+6 y-15=0$

Minimum distance of line from circle

$=\frac{3}{2}-1=\frac{1}{2}$

Standard 11

Mathematics

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