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Question Number 36034 by solihin last updated on 27/May/18

Commented by abdo mathsup 649 cc last updated on 27/May/18

we have z^− (z−i) =10 +2i let put z =x+iy ⇒ (x−iy)(x+iy−i) =10 +2i ⇔ x^2 +ixy −ix −ixy +y^2 −y =10 +2i ⇔ x^2 +y^2 −y + −ix =10 +2i ⇔ x=−2 and x^2 +y^2 −y =10 ⇒ y^2 −y =6 and x=−2 let solve y^2 −y −6 =0 Δ= 1 +24 =25 ⇒ y_1 = ((1 +5)/2) = 3 y_2 = ((1−5)/2) =−2 so z =−2 +3i or z =−2−2i are the solutions for this equation .

wehavez−(z−i)=10+2iletputz=x+iy⇒(x−iy)(x+iy−i)=10+2i⇔x2+ixy−ix−ixy+y2−y=10+2i⇔x2+y2−y+−ix=10+2i⇔x=−2andx2+y2−y=10⇒y2−y=6andx=−2letsolvey2−y−6=0Δ=1+24=25⇒y1=1+52=3y2=1−52=−2soz=−2+3iorz=−2−2iarethesolutionsforthisequation.

Answered by Rasheed.Sindhi last updated on 27/May/18

Let z=x+iy ∴ z′=x−iy z′(z−i)=10+2i (x−iy)(x+iy−i)=10+2i (x−iy)( x+(y−1)i )=10+2i x^2 +x(y−1)i−xyi+y(y−1)=10+2i x^2 +y(y−1)+x{(y−1)−y}i=10+2i x^2 +y(y−1)−xi=10+2i x^2 +y(y−1)=10 ∧ x=−2 (−2)^2 +y^2 −y=10 y^2 −y−10+4=0 y^2 −y−6=0 (y−3)(y+2)=0 y=3 ∣ y=−2 z=x+iy=−2+3i , −2−2i

Letz=x+iy∴z′=x−iyz′(z−i)=10+2i(x−iy)(x+iy−i)=10+2i(x−iy)(x+(y−1)i)=10+2ix2+x(y−1)i−xyi+y(y−1)=10+2ix2+y(y−1)+x{(y−1)−y}i=10+2ix2+y(y−1)−xi=10+2ix2+y(y−1)=10∧x=−2(−2)2+y2−y=10y2−y−10+4=0y2−y−6=0(y−3)(y+2)=0y=3∣y=−2z=x+iy=−2+3i,−2−2i

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