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Question 79 of 122

Q.'P' represents the variable complex number zz. Find the locus of P if Re(z+1z+i)=1\text{Re}\left(\dfrac{z+1}{z+i}\right)=1.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★
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Substitute z=x+iyz=x+iy, rationalize the complex fraction, isolate the real part, set it equal to 1, and simplify to identify the locus.

1. Let z=x+iyz=x+iy. Then:

z+1z+i=(x+1)+iyx+i(y+1)\frac{z+1}{z+i}=\frac{(x+1)+iy}{x+i(y+1)}

2. Rationalize by multiplying numerator and denominator by the conjugate of the denominator, x−i(y+1)x-i(y+1):

[(x+1)+iy] [x−i(y+1)]x2+(y+1)2\frac{[(x+1)+iy]\,[x-i(y+1)]}{x^2+(y+1)^2}

3. Expand the numerator.

(x+1)+iy] [x−i(y+1)]=(x+1)x−i(x+1)(y+1)+iyx−i2y(y+1)(x+1)+iy]\,[x-i(y+1)] = (x+1)x - i(x+1)(y+1) + i y x - i^2 y(y+1)

=[x(x+1)+y(y+1)]+i[xy−(x+1)(y+1)]=\big[x(x+1)+y(y+1)\big] + i\big[xy-(x+1)(y+1)\big]

4. Extract the real part.

Re⁡(z+1z+i)=x(x+1)+y(y+1)x2+(y+1)2\operatorname{Re}\left(\frac{z+1}{z+i}\right)=\frac{x(x+1)+y(y+1)}{x^2+(y+1)^2}

5. Set this equal to 1 (given condition) and clear the denominator.

x(x+1)+y(y+1)=x2+(y+1)2x(x+1)+y(y+1)=x^2+(y+1)^2

x2+x+y2+y=x2+y2+2y+1x^2+x+y^2+y=x^2+y^2+2y+1

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