If $z$ is a complex number such that $z\neq0$ and $\mathrm{Re}(z)=0$, then $\mathrm{Im}(z^2)=0$.

If z is a complex number such that z≠0 and mathrmRe(z)=0, then mathrmIm(z²)=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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If a non‑zero complex number is purely imaginary ( $\mathrm{Re}(z)=0$ ), then its square is a negative real number, so $\mathrm{Im}(z^2)=0$ is always true.

Why this works

The core idea is simple: when you square a purely imaginary number, the result is a real number — and a real number has zero imaginary part. This follows directly from how complex multiplication works.

Let’s recall: any complex number can be written as $z = x + iy$ , where $x = \mathrm{Re}(z)$ and $y = \mathrm{Im}(z)$ . The condition $\mathrm{Re}(z) = 0$ means $x = 0$ , so $z$ lies entirely on the imaginary axis. The only exception is $z = 0$ , which is excluded here.

Now, what happens when we square such a number?

Write $z$ in its simplest form

Since $\mathrm{Re}(z) = 0$ , we have $z = 0 + iy = iy$ , where $y$ is a real number. Because $z \neq 0$ , we know $y \neq 0$ .
Square it

z^2 = (iy)^2 = i^2 y^2 = (-1) y^2 = -y^2

Interpret the result $-y^2$ is a real number — it has no imaginary part. So $\mathrm{Im}(z^2) = 0$ .

That’s the entire reasoning. The statement is always true for any non‑zero purely imaginary $z$ . …

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Q.If zzz is a complex number such that z≠0z\neq0z=0 and Re(z)=0\mathrm{Re}(z)=0Re(z)=0, then Im(z2)=0\mathrm{Im}(z^2)=0Im(z2)=0.

Why this works

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Q.If $z$ is a complex number such that $z\neq0$ and $\mathrm{Re}(z)=0$ , then $\mathrm{Im}(z^2)=0$ .

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