If u+iv = frac{3i}{x+iy+2},then y= | vedclass

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(B) Given,$u+iv = \frac{3i}{(x+2)+iy}$.Taking the modulus on both sides,$|u+iv| = |\frac{3i}{(x+2)+iy}|$.$\sqrt{u^2+v^2} = \frac{3}{\sqrt{(x+2)^2+y^2}

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$\frac{3u}{u^2+v^2}$

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(B) Given,$u+iv = \frac{3i}{(x+2)+iy}$.Taking the modulus on both sides,$|u+iv| = |\frac{3i}{(x+2)+iy}|$.$\sqrt{u^2+v^2} = \frac{3}{\sqrt{(x+2)^2+y^2}}$.Squaring both sides,$u^2+v^2 = \frac{9}{(x+2)^2+y^2}$,which implies $(x+2)^2+y^2 = \frac{9}{u^2+v^2} \dots (i)$.Now,$u+iv = \frac{3i((x+2)-iy)}{(x+2)^2+y^2} = \frac{3y + i(3(x+2))}{(x+2)^2+y^2}$.Comparing the real parts,$u = \frac{3y}{(x+2)^2+y^2}$.Substituting $(x+2)^2+y^2 = \frac{9}{u^2+v^2}$ into the expression for $u$:$u = \frac{3y}{9 / (u^2+v^2)} = \frac{3y(u^2+v^2)}{9} = \frac{y(u^2+v^2)}{3}$.Therefore,$y = \frac{3u}{u^2+v^2}$.

If $u+iv = \frac{3i}{x+iy+2}$,then $y=$

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