We have3+ix2y=x2+y+4ix2+y+4i-3-ix2y=0x2+y-3+i4-x2y=0+0iOn comparing we getx2+y-3=0 1 and 4-x2y=0Consider 4-x2y=0⇒x2y=4⇒x2=4y 2Putting 2 in 1, we get4y+y-3=04+y2-3y=0y2-3y+4=0y=--3±-32-41421 uisng quadratic formulay=3±9-162y=3±i72So, x2=4y⇒ x2=43±i72⇒ x2=83±i7Case I when x2=83+i7×3-i73-i7=83-i79+7=3-i72So, x=±3-i72Similarly try with x2=83-i7

Pls solve this...

Pls solve this... Q.ll Solve for x and y: 3 + ix2y=x2+y+4i.

We have 3 + {ix}^{2} y = x^{2} + y + 4 i x^{2} + y + 4 i - 3 - {ix}^{2} y = 0 (x^{2} + y - 3) + i (4 - x^{2} y) = 0 + 0 i On comparing we get x^{2} + y - 3 = 0 . . . (1) and 4 - x^{2} y = 0 Consider 4 - x^{2} y = 0 \Rightarrow x^{2} y = 4 \Rightarrow x^{2} = \frac{4}{y} . . . (2) Putting (2) in (1), we get \frac{4}{y} + y - 3 = 0 4 + y^{2} - 3 y = 0 y^{2} - 3 y + 4 = 0 y = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (4)}}{2 (1)} [uisng quadratic formula] y = \frac{3 \pm \sqrt{9 - 16}}{2} y = \frac{3 \pm i \sqrt{7}}{2} So, x^{2} = \frac{4}{y} \Rightarrow x^{2} = \frac{4}{\frac{3 \pm i \sqrt{7}}{2}} \Rightarrow x^{2} = \frac{8}{3 \pm i \sqrt{7}} Case I when x^{2} = \frac{8}{3 + i \sqrt{7}} \times \frac{3 - i \sqrt{7}}{3 - i \sqrt{7}} = \frac{8 (3 - i \sqrt{7})}{9 + 7} = \frac{3 - i \sqrt{7}}{2} So, x = \pm \sqrt{\frac{3 - i \sqrt{7}}{2}} Similarly try with x^{2} = \frac{8}{3 - i \sqrt{7}}

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Pls solve this... Pls solve this... Q.ll Solve for x and y: 3 + ix2y=x2+y+4i.

Pls solve this...

Pls solve this... Q.ll Solve for x and y: 3 + ix2y=x2+y+4i.