z=x+iy and the point P represents z in the Ar | vedclass

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(B) Let $z = x + iy$. The expression is $\frac{2z-i}{z+2i} = \frac{2(x+iy)-i}{(x+iy)+2i} = \frac{2x + i(2y-1)}{x + i(y+2)}$. To find the argument,mult

Vedclass · Accepted Answer

$2x^2+2y^2+5x+3y-2=0, (x, y) 
eq (0, -2)$

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(B) Let $z = x + iy$. The expression is $\frac{2z-i}{z+2i} = \frac{2(x+iy)-i}{(x+iy)+2i} = \frac{2x + i(2y-1)}{x + i(y+2)}$. To find the argument,multiply the numerator and denominator by the conjugate of the denominator: $\frac{2x + i(2y-1)}{x + i(y+2)} 	imes \frac{x - i(y+2)}{x - i(y+2)} = \frac{2x^2 + (2y-1)(y+2) + i[x(2y-1) - 2x(y+2)]}{x^2 + (y+2)^2}$. The real part is $R = \frac{2x^2 + 2y^2 + 3y - 2}{x^2 + (y+2)^2}$ and the imaginary part is $I = \frac{-5x}{x^2 + (y+2)^2}$. Given $	ext{Arg}\left(\frac{2z-i}{z+2i}ight) = \frac{\pi}{4}$,we have $	an\left(\frac{\pi}{4}ight) = \frac{I}{R} = 1$. Thus,$I = R$,which implies $\frac{-5x}{x^2 + (y+2)^2} = \frac{2x^2 + 2y^2 + 3y - 2}{x^2 + (y+2)^2}$. This simplifies to $2x^2 + 2y^2 + 5x + 3y - 2 = 0$,where $(x, y) 
eq (0, -2)$.

$z=x+iy$ and the point $P$ represents $z$ in the Argand plane. If the amplitude of $\left(\frac{2z-i}{z+2i}\right)$ is $\frac{\pi}{4}$,then the equation of the locus of $P$ is

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