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Question Number 19629 by Tinkutara last updated on 13/Aug/17

If ∣z∣ = 2, then the points representing  the complex numbers −1 + 5z will lie  on a  (1) Circle  (2) Straight line  (3) Parabola  (4) Ellipse

$$\mathrm{If}\:\mid{z}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:−\mathrm{1}\:+\:\mathrm{5}{z}\:\mathrm{will}\:\mathrm{lie} \\ $$$$\mathrm{on}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Ellipse} \\ $$

Answered by ajfour last updated on 13/Aug/17

x+iy=−1+5×2e^(iθ)   x−iy=−1+5×2e^(−iθ)   (x+1+iy)(x+1−iy)=100  (x+1)^2 +y^2 =10^2    They lie on a circle then.  (1).

$$\mathrm{x}+\mathrm{iy}=−\mathrm{1}+\mathrm{5}×\mathrm{2e}^{\mathrm{i}\theta} \\ $$$$\mathrm{x}−\mathrm{iy}=−\mathrm{1}+\mathrm{5}×\mathrm{2e}^{−\mathrm{i}\theta} \\ $$$$\left(\mathrm{x}+\mathrm{1}+\mathrm{iy}\right)\left(\mathrm{x}+\mathrm{1}−\mathrm{iy}\right)=\mathrm{100} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} \: \\ $$$$\mathrm{They}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{then}.\:\:\left(\mathrm{1}\right). \\ $$

Commented by Tinkutara last updated on 13/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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