Find the complex number satisfying the equation $z+\sqrt{2}\,|(z+1)|+i=0$.

> [!TLDR] > The key idea is to treat the complex number as $z = x + iy$, separate the equation into real and imaginary parts, and solve the resulting system. The solution is $z = -2 - i$. We have an equation mixing $z$ itself, the modulus of $(z+1)$, and a constant $i$. The modulus $|z+1|$ is a real number, so the whole expression $z + \sqrt{2}\,|z+1| + i = 0$ must be split into its real and imaginary parts to be solved. Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $z+1 = (x+1) + iy$, so its modulus is $$ |z+1| = \sqrt{(x+1)^2 + y^2}. $$ Substitute into the given equation: $$ (x + iy) + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} + i = 0. $$ Group the real and imaginary parts: $$ \left[ x + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} \right] + i\,(y + 1) = 0 + 0i. $$ For a complex number to be zero, both its real and imaginary parts must be zero. This gives us two equations: 1. **Imaginary part:** $y + 1 = 0 \implies y = -1$. 2. **Real part:** $x + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} = 0$. Now substitute $y = -1$ into the real part equation: $$ x + \sqrt{2}\,\sqrt{(x+1)^2 + (-1)^2} = 0. $$ Since $(-1)^2 = 1$, we have: $$ x + \sqrt{2}\,\sqrt{(x+1)^2 + 1} = 0. $$ > [!WARNING] > A common mistake is to forget that the modulus $\sqrt{(x+1)^2 + 1}$ is always positive. The equation $x + \sqrt{2}\,\sqrt{(x+1)^2 + 1} = 0$ forces $x$ to be negative, because the square root term is positive. If you try to square both sides without first isolating the square root, you might introduce extraneous solutions. Isolate the square root term: $$ \sqrt{2}\,\sqrt{(x+1)^2 + 1} = -x. $$ Since the left side is non-negative, the right side $-x$ must also be non-negative, so $x \le 0$. This is consistent with our expectation. …

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

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z+\sqrt{2}\,|(z+1)|+i=0

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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The key idea is to treat the complex number as $z = x + iy$ , separate the equation into real and imaginary parts, and solve the resulting system. The solution is $z = -2 - i$ .

We have an equation mixing $z$ itself, the modulus of $(z+1)$ , and a constant $i$ . The modulus $|z+1|$ is a real number, so the whole expression $z + \sqrt{2}\,|z+1| + i = 0$ must be split into its real and imaginary parts to be solved.

Let $z = x + iy$ , where $x, y \in \mathbb{R}$ . Then $z+1 = (x+1) + iy$ , so its modulus is

|z+1| = \sqrt{(x+1)^2 + y^2}.

Substitute into the given equation:

(x + iy) + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} + i = 0.

Group the real and imaginary parts:

\left[ x + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} \right] + i\,(y + 1) = 0 + 0i.

For a complex number to be zero, both its real and imaginary parts must be zero. This gives us two equations:

Imaginary part: $y + 1 = 0 \implies y = -1$ .
Real part: $x + \sqrt{2}\,\sqrt{(x+1)^2 + y^2} = 0$ .

Now substitute $y = -1$ into the real part equation:

x + \sqrt{2}\,\sqrt{(x+1)^2 + (-1)^2} = 0.

Since $(-1)^2 = 1$ , we have:

x + \sqrt{2}\,\sqrt{(x+1)^2 + 1} = 0.

Watch out

A common mistake is to forget that the modulus $\sqrt{(x+1)^2 + 1}$ is always positive. The equation $x + \sqrt{2}\,\sqrt{(x+1)^2 + 1} = 0$ forces $x$ to be negative, because the square root term is positive. If you try to square both sides without first isolating the square root, you might introduce extraneous solutions.

Isolate the square root term:

\sqrt{2}\,\sqrt{(x+1)^2 + 1} = -x.

Since the left side is non-negative, the right side $-x$ must also be non-negative, so $x \le 0$ . This is consistent with our expectation. …

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Q.Find the complex number satisfying the equation z+2 ∣(z+1)∣+i=0z+\sqrt{2}\,|(z+1)|+i=0z+2​∣(z+1)∣+i=0.

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Q.Find the complex number satisfying the equation $z+\sqrt{2}\,|(z+1)|+i=0$ .

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