Question

5. Find the complex number satisfying the equation
z+√2|z+1|+i=0​

Answer 1

Answer

-2 - i satisfy the given equation.

Solution

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Let z = x+ iy. Then.

z + √2|z+1| + i = 0

x+ iy + √2 [√( x+1)² + y²] + i= 0

x + [√2 (x+1)² + y² ] + i(y + 1) = 0 + 0i

Comparing Re (z) with re(z) and Im(z) with im(z)

x +[√2(x+1)² + y²] = 0

y + 1 = 0 so ,

y = -1

[√2(x+1)² + y² ] = -x

Squaring both sides.

[√2[(x+1)² + y²]] ² = x²

2[(x+1)²+y² ]= x²

2[(x²+1 +2x ) +y² ]= x²

putting y = -1

2x² +2 +4x +2(-1)²= x²

2x²-x² +4x +2 +2 = 0

x² + 4x + 4 = 0

(x+2)² = 0

x +2 = 0

x = -2

Hence , z = x +iy = -2 - i

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Answer 2

AnswEr:

Let z = x + iy . Then,

$\qquad \tt \: z + \sqrt{2} \: |z + 1| + i = 0$

$\implies \tt \: x + iy + \sqrt{2} \: |(x + 1) + iy| \: + i = 0 \\ \\ \implies \ \ \tt x + \sqrt{2} \: \sqrt{ {(x + 1)}^{2} + {y}^{2} } + (y + ) \: i = 0 \\ \\ \tt \implies \: x + \sqrt{2 {(x + 1)}^{2} + 2y {}^{2} } = 0 \: and \: (y + 1) = 0 \\ \\ \implies \tt \: x + \sqrt{2 {(x - 1)}^{2} + 2y {}^{2} } = 0 \: and \: y = - 1 \\ \\ \tt \implies \: x + \sqrt{2 {(x + 1)}^{2} + 2 } = 0 \: and \: y = - 1 \\ \\ \tt \implies \sqrt{2 {(x + 1)}^{2} + 2 } = - x \: and \: y = - 1 \\ \\ \tt \implies \: 2(x + 1) {}^{2} + 2 = {x}^{2} \: and \: y = - 1 \\ \\ \tt \implies {x}^{2} + 4x + 4 = 0 \: and \: y = - 1 \\ \\ \tt \implies {(x + 1)}^{2} = 0 \: and \: y = - 1 \\ \\ \tt \implies \: x = - 2 \: and \: y = - 1$

Hence, z = x + iy = -2 -i.