Solve the system of equations $\mathrm{Re}(z^2)=0$, $|z|=2$.

Solve the system of equations mathrmRe(z²)=0, |z|=2.

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\mathrm{Re}(z^2)=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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A complex number with modulus $2$ whose square has zero real part must lie on the axes at distance $2$ from the origin; squaring rotates the argument by $2$ , so we need $\arg(z) = \pm\frac{\pi}{4}, \pm\frac{3\pi}{4}$ . The four solutions are $\boxed{\pm\sqrt{2}(1+i), \pm\sqrt{2}(1-i)}$ .

The constraint $|z|=2$ tells us that $z$ lives on a circle of radius $2$ centered at the origin. The condition $\mathrm{Re}(z^2)=0$ says that when we square $z$ , the result must be purely imaginary (or zero). Squaring a complex number doubles its argument and squares its modulus, so we're looking for points on the circle whose argument, when doubled, lands on the imaginary axis.

Let's write $z = x + iy$ where $x, y \in \mathbb{R}$ .

Translate the modulus condition. From $|z|=2$ , we have

x^2 + y^2 = 4.

Expand $z^2$ and extract its real part. Compute

z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy.

The real part is $x^2 - y^2$ , so the condition $\mathrm{Re}(z^2)=0$ becomes

x^2 - y^2 = 0 \quad \Longrightarrow \quad x^2 = y^2.

Solve the system $x^2 = y^2$ and $x^2 + y^2 = 4$ .

From $x^2 = y^2$ , either $y = x$ or $y = -x$ .
- Case 1: $y = x$ . Substitute into $x^2 + y^2 = 4$ :

x^2 + x^2 = 4 \quad \Longrightarrow \quad 2x^2 = 4 \quad \Longrightarrow \quad x^2 = 2 \quad \Longrightarrow \quad x = \pm\sqrt{2}.

 This gives $z = \sqrt{2} + i\sqrt{2} = \sqrt{2}(1+i)$ and $z = -\sqrt{2} - i\sqrt{2} = -\sqrt{2}(1+i)$.

Case 2: $y = -x$ . Substitute into $x^2 + y^2 = 4$ : …

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Q.Solve the system of equations Re(z2)=0\mathrm{Re}(z^2)=0Re(z2)=0, ∣z∣=2|z|=2∣z∣=2.

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Q.Solve the system of equations $\mathrm{Re}(z^2)=0$ , $|z|=2$ .

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