The locus of z = x+iy such that {Im}≤ft(z-3i/iz+4) = 0 is

Correct Answer is : x2+y2−7y+12x²+y²-7y+12x2+y2−7y+12 = 000 and (x,y)≠(0,4)(x,y)≠(0,4)(x,y)=(0,4)

x²+y²-7y+12 = 0 and (x,y)≠(0,4)

x²-y²+7y-12 = 0 and (x,y)≠(0,4)

Correct Answer is : x2+y2−7y+12x²+y²-7y+12x2+y2−7y+12 = 000 and (x,y)≠(0,4)(x,y)≠(0,4)(x,y)=(0,4)

Test Index

TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 7 of 160

Marks: +1, -0

The locus of

z

x+iy

\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)

0

Solution:

\; \text{ } \; x^{2}+y^{2}-7y+12=0 \; \text{ and } \; (x, y)\neq(0,4)

\; \text{ Given, } \; z=x+iy \; \text{ and } \; \operatorname{Im}\left(\; \frac{z-3i}{iz+4}\right)=0

\Rightarrow \operatorname{Im}\left[\; \frac{x+iy-3i}{i(x+iy)+4}\right]=0 \Rightarrow \operatorname{Im}\left[\; \frac{x+i(y-3)}{(4-y)+ix}\right]=0

\Rightarrow \operatorname{Im}\left[\; \frac{x+i(y-3)}{(4-y)+ix} \times \; \frac{(4-y)-ix}{(4-y)-ix}\right]=0

\Rightarrow \operatorname{Im}\left[\; \frac{x(4-y)-ix^{2}+i(y-3)(4-y)+x(y-3)}{(4-y)^{2}+x^{2}}\right]=0

\Rightarrow \operatorname{Im}\left[\; \frac{4x-xy-ix^{2}+i(-y^{2}+7y-12)+xy-3x}{(4-y)^{2}+x^{2}}\right]=0

\Rightarrow \operatorname{Im}\left[\; \frac{x-i(x^{2}+y^{2}-7y+12)}{(4-y)^{2}+x^{2}}\right]=0

\Rightarrow \; \frac{x^{2}+y^{2}-7y+12}{(4-y)^{2}+x^{2}}=0

\Rightarrow 4-y \neq 0 \; \text{ and } \; x \neq 0 \Rightarrow y \neq 4 \; \text{ and } \; x \neq 0

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