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Question Number 44094 by peter frank last updated on 21/Sep/18

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Sep/18

is it z^5    or z^3    pls clarify

$${is}\:{it}\:{z}^{\mathrm{5}} \:\:\:{or}\:{z}^{\mathrm{3}} \:\:\:{pls}\:{clarify} \\ $$

Commented by peter frank last updated on 23/Sep/18

z^5

$${z}^{\mathrm{5}} \: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Sep/18

a)((x+i(x−2))/(3+i))+((2y+i(1−3y))/(3−i))=i  [(3−i){x+i(x−2)}+(3+i){2y+i(1−3y)}]/10=i  ((3x+i(3x−6)−ix+(x−2)+6y+i(3−9y)+i2y−(1−3y))/(10))=i  ((4x−2+6y−1+3y)/(10))+i((3x−6−x+3−9y+2y)/(10))=i  ((4x+9y−3)/(10))+i((2x−7y−3)/(10))=i  4x+9y−3=0  2x−7y−3=10  4x+9y−3=0  4x−14y−26=0  23y+23=0   y=−1  4x−9−3=0    x=3

$$\left.{a}\right)\frac{{x}+{i}\left({x}−\mathrm{2}\right)}{\mathrm{3}+{i}}+\frac{\mathrm{2}{y}+{i}\left(\mathrm{1}−\mathrm{3}{y}\right)}{\mathrm{3}−{i}}={i} \\ $$$$\left[\left(\mathrm{3}−{i}\right)\left\{{x}+{i}\left({x}−\mathrm{2}\right)\right\}+\left(\mathrm{3}+{i}\right)\left\{\mathrm{2}{y}+{i}\left(\mathrm{1}−\mathrm{3}{y}\right)\right\}\right]/\mathrm{10}={i} \\ $$$$\frac{\mathrm{3}{x}+{i}\left(\mathrm{3}{x}−\mathrm{6}\right)−{ix}+\left({x}−\mathrm{2}\right)+\mathrm{6}{y}+{i}\left(\mathrm{3}−\mathrm{9}{y}\right)+{i}\mathrm{2}{y}−\left(\mathrm{1}−\mathrm{3}{y}\right)}{\mathrm{10}}={i} \\ $$$$\frac{\mathrm{4}{x}−\mathrm{2}+\mathrm{6}{y}−\mathrm{1}+\mathrm{3}{y}}{\mathrm{10}}+{i}\frac{\mathrm{3}{x}−\mathrm{6}−{x}+\mathrm{3}−\mathrm{9}{y}+\mathrm{2}{y}}{\mathrm{10}}={i} \\ $$$$\frac{\mathrm{4}{x}+\mathrm{9}{y}−\mathrm{3}}{\mathrm{10}}+{i}\frac{\mathrm{2}{x}−\mathrm{7}{y}−\mathrm{3}}{\mathrm{10}}={i} \\ $$$$\mathrm{4}{x}+\mathrm{9}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{x}−\mathrm{7}{y}−\mathrm{3}=\mathrm{10} \\ $$$$\mathrm{4}{x}+\mathrm{9}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{4}{x}−\mathrm{14}{y}−\mathrm{26}=\mathrm{0} \\ $$$$\mathrm{23}{y}+\mathrm{23}=\mathrm{0}\:\:\:{y}=−\mathrm{1} \\ $$$$\mathrm{4}{x}−\mathrm{9}−\mathrm{3}=\mathrm{0}\:\:\:\:{x}=\mathrm{3} \\ $$

Commented by peter frank last updated on 21/Sep/18

thanks

$${thanks} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Sep/18

d)2+i+2−i+α=((−(−6))/1)  4+α=6   α=2  (2+i)(2−i)α=−k  5α=−k  k=−5(2)  k=−10  others roots sre  2+i   2−i  and  2

$$\left.{d}\right)\mathrm{2}+{i}+\mathrm{2}−{i}+\alpha=\frac{−\left(−\mathrm{6}\right)}{\mathrm{1}} \\ $$$$\mathrm{4}+\alpha=\mathrm{6}\:\:\:\alpha=\mathrm{2} \\ $$$$\left(\mathrm{2}+{i}\right)\left(\mathrm{2}−{i}\right)\alpha=−{k} \\ $$$$\mathrm{5}\alpha=−{k}\:\:{k}=−\mathrm{5}\left(\mathrm{2}\right) \\ $$$${k}=−\mathrm{10} \\ $$$${others}\:{roots}\:{sre}\:\:\mathrm{2}+{i}\:\:\:\mathrm{2}−{i}\:\:{and}\:\:\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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