Question

|z+1|=z+2(1+i) find z

Answer 1

Answer:

$z=\frac{1}{2}-2i$

Step-by-step explanation:

Let z = x + iy,

z + 1 = x + iy + 1 = ( x + 1 ) + iy

$\implies |z+1| = \sqrt{(x+1)^2 + y^2}$

Now,

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

$|z+1|=z+2(1+i)$

$\sqrt{(x+1)^2 + y^2} = (x + 2) + (y+2)i$

Equating real parts,

$\sqrt{(x+1)^2 + y^2} = x+2$,

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

$x^2 + 2x + 1 + y^2 = x^2 + 4x + 4$

$2x + 1 + y^2 = 4x + 4$

$y^2 -2x - 3=0-----(1)$

Equating imaginary parts,

$0=y+2\implies y = -2-----(2)$

From equation (1),

$4 -2x - 3 = 0$

$-2x = -1\implies x = \frac{1}{2}$

Hence,

$z=\frac{1}{2}-2i$

Answer 2

Answer:

Refer the above attachment

Step-by-step explanation:

Hope it helps uhh....

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