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If $z+ \sqrt2 |z+1| +i = 0, \text{ find } |z|$

My attempt:

As the RHS $= 0$, the sum of the real parts and imaginary parts are both $0$.

As the amplitude of a complex number is always real, $z + i = 0 \iff z=i$. Moreover, $|i+1|=0$. Thus, $|z|=1$.

However, my book says $|z| = \sqrt{5}$. Have I made a mistake anywhere?

Eric Wofsey
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math and physics forever
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  • This is a wrong argument. Yes, it is true that $|z+1|$ is non-negative and real, but that does not imply $z+i$ is zero. That means the (real part of $z$)+$\sqrt{2}|z+i|$ is zero and (imaginary part of $z$)+$1$ is zero. – ShyamalSayak Sep 16 '23 at 13:02
  • Hint: $\frac{|z-z_0|}{|z-z_1|}=cst$ is a circle, find points on that circle that fits the original equation. – zwim Sep 16 '23 at 15:10

2 Answers2

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Stating $z + i = 0$ is false. This follows from $z$ being complex (having a real and imaginary), not purely imaginary. Only $\Im{(z)} + i = 0$.

If we assume $z = a + bi$, then we can correctly say $bi + i = 0$ and $a = -\sqrt{2}|z + 1|$. Hence, $(b + 1) = 0 \iff b = -1$.

After substitution,

$$a = -\sqrt{2}|z + 1| = -\sqrt{2}\cdot\sqrt{a^2 + 2a + b^2 + 1} \\ a^2 = 2a^2 + 4a + 4 \iff (a + 2)^2 = 0 \implies a = -2$$

Then, $|z| = \sqrt{a^2 + b^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5}$

Dstarred
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Given $\displaystyle z+\sqrt{2}|z+1|+i=0$

$\displaystyle z+i=-\sqrt{2}|z+1|$ Purely real number.

So $z=a-i,$ Where $a<0$

So $\displaystyle a=-\sqrt{2}\sqrt{(a+1)^2+1^2}$

$\displaystyle a^2=2[(a+1)^2+1]$

$a^2=2[a^2+2a+2]\Longrightarrow a^2+4a+4=0$

So we get $a=-2$ and

$z=-2-i\Longrightarrow |z|=\sqrt{5}$

jacky
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