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NCERT Exemplar · Q16

Q.If ∣z+1∣=z+2(1+i)|z+1|=z+2(1+i), then find zz.

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The right side must be real, forcing Im⁡(z)=−2\operatorname{Im}(z)=-2; solving the resulting real equation gives z=12−2iz = \tfrac{1}{2} - 2i.

Let z=x+iyz = x+iy. The left side ∣z+1∣=(x+1)2+y2|z+1| = \sqrt{(x+1)^2+y^2} is real and non-negative, so the right side must be real:

z+2(1+i)=(x+2)+i(y+2).z + 2(1+i) = (x+2) + i(y+2).

Its imaginary part must vanish:

y+2=0  ⟹  y=−2.y + 2 = 0 \implies y = -2.

Substituting y=−2y=-2 (and requiring x+2≥0x+2 \ge 0):

(x+1)2+4=x+2.\sqrt{(x+1)^2 + 4} = x + 2.

Squaring:

(x+1)2+4=(x+2)2  ⟹  x2+2x+5=x2+4x+4  ⟹  2x=1  ⟹  x=12.(x+1)^2 + 4 = (x+2)^2 \implies x^2 + 2x + 5 = x^2 + 4x + 4 \implies 2x = 1 \implies x = \tfrac{1}{2}. …

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