The locus of a point on the Argand plane repr | vedclass

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(A) Given condition is $\left|\frac{z-1+i}{z+1-i}ight|=\left|\operatorname{Re}\left(\frac{z-1+i}{z+1-i}ight)ight|$, where $z 
eq -1+i$. Let $w

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$A$ straight line that does not contain the point $(-1+i)$

Vedclass · Answer

(A) Given condition is $\left|\frac{z-1+i}{z+1-i}ight|=\left|\operatorname{Re}\left(\frac{z-1+i}{z+1-i}ight)ight|$, where $z 
eq -1+i$. Let $w = \frac{z-1+i}{z+1-i}$. The condition is $|w| = |\operatorname{Re}(w)|$. For any complex number $w = u + iv$, $|w| = \sqrt{u^2 + v^2}$ and $|\operatorname{Re}(w)| = |u| = \sqrt{u^2}$. Thus, $\sqrt{u^2 + v^2} = \sqrt{u^2} \implies u^2 + v^2 = u^2 \implies v^2 = 0 \implies v = 0$. This means $\operatorname{Im}(w) = 0$, so $w$ must be a purely real number. Let $z = x + iy$. Then $w = \frac{(x-1) + i(y+1)}{(x+1) + i(y-1)}$. Multiplying numerator and denominator by the conjugate of the denominator: $w = \frac{((x-1) + i(y+1))((x+1) - i(y-1))}{(x+1)^2 + (y-1)^2}$. The imaginary part of $w$ is zero when the imaginary part of the numerator is zero: $(x-1)(-(y-1)) + (y+1)(x+1) = 0$. $-(xy - x - y + 1) + (xy + x + y + 1) = 0$. $-xy + x + y - 1 + xy + x + y + 1 = 0$. $2x + 2y = 0 \implies x + y = 0$. Since $z 
eq -1+i$, the point $(-1, 1)$ is excluded from the line $x+y=0$. Thus, the locus is a straight line that does not contain the point $(-1+i)$.

The locus of a point on the Argand plane represented by the complex number $z$, when $z$ satisfies the condition $\left|\frac{z-1+i}{z+1-i}\right|=\left|\operatorname{Re}\left(\frac{z-1+i}{z+1-i}\right)\right|$ is

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