Which of the following is correct for any two complex numbers $z_1$ and $z_2$? (A) $|z_1z_2|=|z_1||z_2|$ (B) $\arg(z_1z_2)=\arg(z_1)\cdot\arg(z_2)$ (C) $|z_1+z_2|=|z_1|+|z_2|$ (D) $|z_1+z_2|\geq|z_1|-|z_2|$

> [!TLDR] > Testing each option against the modulus properties shows that **(A)** $|z_1z_2|=|z_1||z_2|$ and **(D)** $|z_1+z_2|\geq|z_1|-|z_2|$ both hold for any two complex numbers — this NCERT Exemplar question has two correct options. This question checks whether you know how the modulus behaves under multiplication and addition of complex numbers. Let's examine each option in turn. ## Checking each option 1. **Option (A): $|z_1z_2|=|z_1||z_2|$** Write $z_1=a+bi$, $z_2=c+di$. The modulus is multiplicative for complex numbers: $|z_1z_2|^2=|z_1|^2|z_2|^2$ holds identically, since $|z_1z_2|^2=(z_1z_2)\overline{(z_1z_2)}=z_1\overline{z_1}\cdot z_2\overline{z_2}=|z_1|^2|z_2|^2$ for every pair of complex numbers, with no exceptions. **(A) is true.** 2. **Option (B): $\arg(z_1z_2)=\arg(z_1)\cdot\arg(z_2)$** Arguments of a product **add**, they do not multiply — this is a standard property of complex numbers. So $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ (up to a multiple of $2\pi$), not a product. **(B) is false.** 3. **Option (C): $|z_1+z_2|=|z_1|+|z_2|$** This is the **triangle inequality**, and it only holds as an *equality* in the special case where $z_1$ and $z_2$ point in the same direction from the origin. For two arbitrary complex numbers the correct general statement is $|z_1+z_2|\le|z_1|+|z_2|$, so equality is not guaranteed for *any* two complex numbers. **(C) is false.** …

Which of the following is correct for any two complex numbers z₁ and z₂? (A) |z₁z₂|=|z₁||z₂| (B) arg(z₁z₂)=arg

NCERT Exemplar · Q29

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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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Testing each option against the modulus properties shows that (A) $|z_1z_2|=|z_1||z_2|$ and (D) $|z_1+z_2|\geq|z_1|-|z_2|$ both hold for any two complex numbers — this NCERT Exemplar question has two correct options.

This question checks whether you know how the modulus behaves under multiplication and addition of complex numbers. Let's examine each option in turn.

Checking each option

Option (A): $|z_1z_2|=|z_1||z_2|$

Write $z_1=a+bi$ , $z_2=c+di$ . The modulus is multiplicative for complex numbers: $|z_1z_2|^2=|z_1|^2|z_2|^2$ holds identically, since $|z_1z_2|^2=(z_1z_2)\overline{(z_1z_2)}=z_1\overline{z_1}\cdot z_2\overline{z_2}=|z_1|^2|z_2|^2$ for every pair of complex numbers, with no exceptions. (A) is true.
Option (B): $\arg(z_1z_2)=\arg(z_1)\cdot\arg(z_2)$

Arguments of a product add, they do not multiply — this is a standard property of complex numbers. So $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ (up to a multiple of $2\pi$ ), not a product. (B) is false.
Option (C): $|z_1+z_2|=|z_1|+|z_2|$

This is the triangle inequality, and it only holds as an equality in the special case where $z_1$ and $z_2$ point in the same direction from the origin. For two arbitrary complex numbers the correct general statement is $|z_1+z_2|\le|z_1|+|z_2|$ , so equality is not guaranteed for any two complex numbers. (C) is false. …

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Checking each option

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