Let x, y∈mathbfR, then x+iy is a non real complex number if: (A) x=0 (B) y=0 (C) x≠0 (D) y≠0

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x, y\in\mathbf{R}

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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 $x + iy$ is non-real precisely when its imaginary part is non-zero; the answer is (D) $y \neq 0$ .

A complex number $z = x + iy$ (where $x, y \in \mathbb{R}$ ) lives in the complex plane with $x$ as the real part and $y$ as the imaginary part. The question asks when this number is non-real, which means it cannot be represented as a point on the real axis alone.

The real numbers sit inside the complex numbers as the special case where the imaginary part vanishes. A complex number is real if and only if $y = 0$ , because then $z = x + i \cdot 0 = x$ , which is just a real number. Conversely, $z$ is non-real when it has a genuine imaginary component—when it steps off the real axis.

Let me examine each option:

Option (A): $x = 0$

If $x = 0$ , then $z = 0 + iy = iy$ . This is a purely imaginary number (assuming $y \neq 0$ ). While purely imaginary numbers are indeed non-real, this condition is too restrictive: it excludes numbers like $2 + 3i$ , which are also non-real but have $x \neq 0$ . So (A) gives a sufficient but not necessary condition.
Option (B): $y = 0$

If $y = 0$ , then $z = x + i \cdot 0 = x$ , a real number. This is the opposite of what we want—it guarantees $z$ is real, not non-real.
Option (C): $x \neq 0$ …

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Q.Let x,y∈Rx, y\in\mathbf{R}x,y∈R, then x+iyx+iyx+iy is a non real complex number if: (A) x=0x=0x=0 (B) y=0y=0y=0 (C) x≠0x\neq0x=0 (D) y≠0y\neq0y=0

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Q.Let $x, y\in\mathbf{R}$ , then $x+iy$ is a non real complex number if:
(A) $x=0$
(B) $y=0$
(C) $x\neq0$
(D) $y\neq0$

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