Locus of complex number satisfying - z -5 - 4iz-3 -2i - - - -4 is the arc of a circle

z_0 ( 3 + 2i)z_0 (5 4i) = BDAD e^iπ/2 = i ⇒ z_0 + 3 2i = iz_0 5i 4 ⇒ z_0 = 2 5i ⇒ Radius AD =|5 4i ( 2 5i)| = |7 + i| = √(50) = 5√(2) units Length of ...

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Byju's Answer

Standard XI

Mathematics

Equation of Circle in Complex Form

Locus of comp...

Question

Locus of complex number satisfying $arg [\frac{(z - 5 + 4 i)}{(z + 3 - 2 i)}] = - \frac{π}{4}$ is the arc of a circle

whose radius is

5 \sqrt{2}

units

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whose radius is

5

units

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whose length (of arc) is

\frac{15 π}{\sqrt{2}}

units

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whose centre is

- 2 - 5 i

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Solution

The correct option is D whose centre is $- 2 - 5 i$

$\frac{z_{0} - (- 3 + 2 i)}{z_{0} - (5 - 4 i)} = \frac{B D}{A D} e^{i π / 2} = i$
$\Rightarrow z_{0} + 3 - 2 i = i z_{0} - 5 i - 4$
$\Rightarrow z_{0} = - 2 - 5 i$
$\Rightarrow$ Radius $A D = | 5 - 4 i - (- 2 - 5 i) |$
$= | 7 + i |$
$= \sqrt{50} = 5 \sqrt{2}$ units
Length of arc $= \frac{3}{4}$ (perimeter of circle)
$= \frac{3}{4} (2 π \times 5 \sqrt{2})$
$= \frac{15 π}{\sqrt{2}}$ units

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arg [z - (1 + i)] = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{3 π}{4} & ; | z | \leq | z - 2 | - \frac{π}{4} & ; | z | > | z - 2 | \end{matrix}

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Locus of complex number satisfying arg[(z−5+4i)(z+3−2i)]=−π4 is the arc of a circle

The correct option is D whose centre is −2−5i z0−(−3+2i)z0−(5−4i)=BDADeiπ/2=i ⇒z0+3−2i=iz0−5i−4 ⇒z0=−2−5i ⇒ Radius AD=|5−4i−(−2−5i)| =|7+i| =√50=5√2 units Length of arc =34 (perimeter of circle) =34(2π×5√2) =15π√2 units

Locus of complex number satisfying $arg [\frac{(z - 5 + 4 i)}{(z + 3 - 2 i)}] = - \frac{π}{4}$ is the arc of a circle