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If-the-imaginary-part-of-2z-1-iz-1-is-2-then-the-locus-of-the-point-representing-z-in-the-complex-plane-is-

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Question Number 19734 by Tinkutara last updated on 15/Aug/17
If the imaginary part of ((2z + 1)/(iz + 1)) is −2, then the locus of the point representing z in the complex plane is
$$\mathrm{If}\:\mathrm{the}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\frac{\mathrm{2}{z}\:+\:\mathrm{1}}{{iz}\:+\:\mathrm{1}}\:\mathrm{is}\:−\mathrm{2}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{representing} \\ $$$${z}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 15/Aug/17
((2z+1)/(iz+1))=(((2z+1)(1−iz^� ))/((iz+1)(1−iz^� ))) =((2z−2i∣z^2 ∣+1−iz^� )/(iz+∣z∣^2 +1−iz^� )) if z=x+iy =((2(x+iy)−2i(x^2 +y^2 )+1−ix−y)/(1+x^2 +y^2 −2y)) as imaginary part of above complex number is =−2 , we can write ((2y−2x^2 −2y^2 −x)/(1+x^2 +y^2 −2y))=−2 or 2y+x−2=0 ⇒ 4iy+2ix−4i=0 or 2(z−z^� )+i(z+z^� )−4i=0 (2+i)z−(2−i)z^� −4i=0 ⇒ (1−2i)z+(1+2i)z^� +4=0 .
$$\:\:\frac{\mathrm{2z}+\mathrm{1}}{\mathrm{iz}+\mathrm{1}}=\frac{\left(\mathrm{2z}+\mathrm{1}\right)\left(\mathrm{1}−\mathrm{i}\bar {\mathrm{z}}\right)}{\left(\mathrm{iz}+\mathrm{1}\right)\left(\mathrm{1}−\mathrm{i}\bar {\mathrm{z}}\right)} \\ $$$$\:\:=\frac{\mathrm{2z}−\mathrm{2i}\mid\mathrm{z}^{\mathrm{2}} \mid+\mathrm{1}−\mathrm{i}\bar {\mathrm{z}}}{\mathrm{iz}+\mid\mathrm{z}\mid^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\bar {\mathrm{z}}} \\ $$$$\mathrm{if}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\:\:=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{iy}\right)−\mathrm{2i}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)+\mathrm{1}−\mathrm{ix}−\mathrm{y}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2y}} \\ $$$$\mathrm{as}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\mathrm{above}\:\mathrm{complex} \\ $$$$\mathrm{number}\:\mathrm{is}\:=−\mathrm{2}\:,\:\mathrm{we}\:\mathrm{can}\:\mathrm{write} \\ $$$$\:\:\frac{\mathrm{2y}−\mathrm{2x}^{\mathrm{2}} −\mathrm{2y}^{\mathrm{2}} −\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2y}}=−\mathrm{2} \\ $$$$\mathrm{or}\:\:\:\:\mathrm{2y}+\mathrm{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{4iy}+\mathrm{2ix}−\mathrm{4i}=\mathrm{0} \\ $$$$\:\mathrm{or}\:\:\:\:\:\mathrm{2}\left(\mathrm{z}−\bar {\mathrm{z}}\right)+\mathrm{i}\left(\mathrm{z}+\bar {\mathrm{z}}\right)−\mathrm{4i}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left(\mathrm{2}+\mathrm{i}\right)\mathrm{z}−\left(\mathrm{2}−\mathrm{i}\right)\bar {\mathrm{z}}−\mathrm{4i}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{1}−\mathrm{2i}\right)\mathrm{z}+\left(\mathrm{1}+\mathrm{2i}\right)\bar {\mathrm{z}}+\mathrm{4}=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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