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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

isitz5orz3plsclarify

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

z^5

z5

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

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=i3x+i(3x−6)−ix+(x−2)+6y+i(3−9y)+i2y−(1−3y)10=i4x−2+6y−1+3y10+i3x−6−x+3−9y+2y10=i4x+9y−310+i2x−7y−310=i4x+9y−3=02x−7y−3=104x+9y−3=04x−14y−26=023y+23=0y=−14x−9−3=0x=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

d)2+i+2−i+α=−(−6)14+α=6α=2(2+i)(2−i)α=−k5α=−kk=−5(2)k=−10othersrootssre2+i2−iand2

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