If a complex number z satisfies the equation | vedclass

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(C) Given equation is $z + \sqrt{2} |z + 1| + i = 0$.Let $z = x + iy$. Substituting this into the equation:$(x + iy) + \sqrt{2} |x + iy + 1| + i = 0$$

Vedclass · Accepted Answer

$\sqrt{5}$

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(C) Given equation is $z + \sqrt{2} |z + 1| + i = 0$.Let $z = x + iy$. Substituting this into the equation:$(x + iy) + \sqrt{2} |x + iy + 1| + i = 0$$(x + iy) + \sqrt{2} \sqrt{(x + 1)^2 + y^2} + i = 0$Equating the real and imaginary parts to zero:Real part: $x + \sqrt{2} \sqrt{(x + 1)^2 + y^2} = 0$Imaginary part: $y + 1 = 0 \Rightarrow y = -1$Substitute $y = -1$ into the real part equation:$x + \sqrt{2} \sqrt{(x + 1)^2 + (-1)^2} = 0$$x + \sqrt{2} \sqrt{x^2 + 2x + 1 + 1} = 0$$\sqrt{2} \sqrt{x^2 + 2x + 2} = -x$Squaring both sides:$2(x^2 + 2x + 2) = x^2$$2x^2 + 4x + 4 = x^2$$x^2 + 4x + 4 = 0$$(x + 2)^2 = 0 \Rightarrow x = -2$Thus,$z = -2 - i$. The modulus $|z| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.

If a complex number $z$ satisfies the equation $z + \sqrt{2} |z + 1| + i = 0$,then $|z|$ is equal to

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