Question

solve the equation |Z+1| = Z+2(1+i)

Answer 1

|z+1|=|x+iy+1|=|x+1+iy|=(x+1)^2+1=x^2+1+2x+1
=x^2+2x+2
z+2(1+i)=x+iy+2+2i=x+2+i(y+2)
so
x^2+x=i(y+2)
now try to solve next by yourself frnd.
pls mark it as brainiest if it helped

Answer 2

AnswEr:

Let z = x + i y . Then, z + 1 = (x+1) +iy

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$\therefore \: </p><p>\blue{\boxed{\underline{\pink{\sf{ |z + 1| = \sqrt{ {(x + 1)}^{2} + {y}^{2} } }}}}}$

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Now, |z+1| = z + 2 (1+i)

$\implies \tt \sqrt{ {(x + 1)}^{2} + {y}^{2} } = (x + iy) + 2(1 + i) \\ \\ \implies \tt \: \sqrt{ {(x + 1)}^{2} + {y}^{2} } + 0i = (x + 2) \\ + \tt(y + 2)i \\ \\ \implies \tt \: \sqrt{ {(x + 1)}^{2} + {y}^{2} } = x + 2 \: \: and \: \: y + 2 = 0 \\ \\ \implies \tt \: {(x + 1)}^{2} + {y}^{2} = {(x + 2)}^{2} \: \: and \: \: y = - 2 \\ \\ \implies \tt \: {y}^{2} = 2x + 3 \: \: \: and \: \: \: y = - 2 \\ \\ \implies \tt \: x = \frac{1}{2} \: \: \: and \: \: \: y = - 2 \\ \\ \\ </p><p></p><p>\green{\boxed{\pink{\underline{\red{\mathrm{hence \: \: z = \frac{1}{2} - 2i }}}}}}$