1.

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

Answer»

Answer:

<klux>Z</klux>=\frac{1}{2}-<klux>2I</klux>

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


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