Solve the equation |z|=z+1+2i.

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|z|=z+1+2i

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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 write $z$ in the form $x+iy$ , equate the modulus to the given expression, and then compare real and imaginary parts. The only solution is $z = \frac{3}{2} - 2i$ .

When you see an equation mixing $|z|$ (a real, non-negative number) with $z$ itself (a complex number), the natural move is to separate the real and imaginary parts. The modulus $|z|$ is always real, so the right-hand side $z+1+2i$ must also be real. That’s the first big clue — it forces the imaginary part of $z$ to cancel the $2i$ term.

Let’s work through it.

Write $z$ in Cartesian form. Let $z = x + iy$ , where $x, y \in \mathbb{R}$ . Then $|z| = \sqrt{x^2 + y^2}$ . The equation becomes:

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

The left side is real; so the right side must be real. That means its imaginary part must be zero:

y + 2 = 0 \quad \Rightarrow \quad y = -2.

Watch out

A common mistake is to forget that $|z|$ is always real. If you don’t set the imaginary part to zero first, you’ll end up with a messy system. Always check: the modulus is a real number, so whatever it equals must also be real.

Now substitute $y = -2$ into the equation. The modulus becomes $\sqrt{x^2 + (-2)^2} = \sqrt{x^2 + 4}$ . The right side becomes $(x+1) + i(0) = x+1$ . So we have:

\sqrt{x^2 + 4} = x + 1.

Solve for $x$ . Since the left side is non-negative, the right side must also be non-negative: $x+1 \ge 0$ , i.e. $x \ge -1$ . …

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Q.Solve the equation ∣z∣=z+1+2i|z|=z+1+2i∣z∣=z+1+2i.

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Q.Solve the equation $|z|=z+1+2i$ .

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