The value of $\sqrt{-25}\times\sqrt{-9}$ is _____.

The value of √(-25)×√(-9) is ___.

NCERT Exemplar · Q39

\sqrt{-25}\times\sqrt{-9}

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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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When multiplying square roots of negative numbers, convert each to imaginary form first: $\sqrt{-25} \times \sqrt{-9} = 5i \times 3i = 15i^2 = -15$ .

The trap here is tempting: you might want to write $\sqrt{-25} \times \sqrt{-9} = \sqrt{(-25)(-9)} = \sqrt{225} = 15$ . That would be wrong. The rule $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ holds only when at least one of $a$ or $b$ is non-negative. Once both are negative, we've left the real numbers and entered the complex plane, where the algebra of square roots changes.

The correct approach is to recognize that the square root of a negative number is an imaginary number. Recall that $i = \sqrt{-1}$ , so any $\sqrt{-k}$ for positive $k$ can be written as $\sqrt{k} \cdot i$ .

Step-by-step solution

Convert each square root to imaginary form.

\sqrt{-25} = \sqrt{25 \cdot (-1)} = \sqrt{25} \cdot \sqrt{-1} = 5i

$\sqrt{-9} = \sqrt{9 \cdot (-1)} = \sqrt{9} \cdot \sqrt{-1} = 3i$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 94 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\sqrt{5} + 3i$
(a) $\sqrt{5} - 3i$
(b) $\dfrac{\sqrt{5}}{14} + i \cdot \dfrac{3}{14}$
(c) $\dfrac{\sqrt{5}}{14} - i \cdot \dfrac{3}{14}$
(d) $3 - \sqrt{5}i$
›Reveal solutionSolution

The multiplicative inverse of $a+ib$ is $\dfrac{a-ib}{a^2+b^2}$ ; here $a=\sqrt5, b=3$ .

The multiplicative inverse of $z = \sqrt5 + 3i$ is $\dfrac{1}{z}$ . Rationalise by multiplying by the conjugate $\sqrt5 - 3i$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{1+i}{1-i} - \dfrac{1-i}{1+i}$
(a) 2
(b) -2
(c) 1
(d) -1
›Reveal solutionSolution

Simplify each fraction using the conjugate, subtract, then take the modulus of the resulting purely imaginary number.

$\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{(1-i)(1+i)} = \dfrac{1+2i+i^2}{1-i^2} = \dfrac{1+2i-1}{1+1} = \dfrac{2i}{2} = i$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\left(\dfrac{1+i}{1-i}\right)^m = 1$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

Simplify the base to $i$ , then find the smallest positive integer power of $i$ equal to 1 (the order of $i$ is 4).

As shown above, $\dfrac{1+i}{1-i} = i$ . So the equation becomes $i^m = 1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$i^{28}$
(a) $i$
(b) $-i$
(c) $1$
(d) $-1$
›Reveal solutionSolution

$i^{28} = 1$ , option (c).

Recall $i^2=-1$ , so $i^4 = (i^2)^2 = (-1)^2 = 1$ . Powers of $i$ cycle with period 4: $i, -1, -i, 1, i, -1, \dots$

…

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CBSE 2026Set ANNUAL1 mark
$2 - 3i$
›Reveal solutionSolution

The conjugate of $2-3i$ is $2+3i$ .

For a complex number $z=a+ib$ , the conjugate $\bar z = a - ib$ — the sign of the imaginary part is reversed while the real part stays unchanged.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Multiplicative inverse of 1 + \sqrt3i is:$
(a) 1 - √3i
(b) (1 - √3i)/2
(c) (1 + √3i)/4
(d) None of these
›Reveal solutionSolution

z⁻¹ = conjugate(z)/|z|²; for z = 1+√3i this gives (1−√3i)/4, which isn't among options (a)-(c) as written, so the correct choice is (d).

For a complex number $z = 1 + \sqrt{3}i$ , the multiplicative inverse is:

$z^{-1} = \frac{\bar z}{|z|^2}$

Here $\bar z = 1 - \sqrt{3}i$ , and $|z|^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4$ .

So:

$z^{-1} = \frac{1-\sqrt3 i}{4}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. The modulus of (1 + i)/(1 - i) is ..............$
›Reveal solutionSolution

Rationalize (1+i)/(1-i) to a+bi form, or use |z1/z2| = |z1|/|z2| — both give modulus 1.

Method using the modulus quotient rule $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ :

$|1+i| = \sqrt{1^2+1^2}=\sqrt2, \qquad |1-i|=\sqrt{1^2+(-1)^2}=\sqrt2$

$\left|\frac{1+i}{1-i}\right| = \frac{\sqrt2}{\sqrt2} = 1$

…

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CBSE 2026Set ANNUAL1 markMCQ
$i^9 \cdot i^{19}$
(a) $-1$
(b) $-i$
(c) $1$
(d) $0$
›Reveal solutionSolution

Using $i^4=1$ , reduce the exponents mod 4: $i^9 = i$ and $i^{19}=i^3=-i$ , so the product is $i\cdot(-i) = 1$ .

Recall $i^2=-1$ , and powers of $i$ repeat with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ .

$i^9 = i^{4\times2+1} = (i^4)^2 \cdot i = 1\cdot i = i$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$3 + 5i$
(a) $-3+5i$
(b) $-3-5i$
(c) $3-5i$
(d) $3+i$
›Reveal solutionSolution

The conjugate of $z=a+ib$ is $\bar{z}=a-ib$ ; only the sign of the imaginary part changes.

Here $z = 3+5i$ , so $a=3$ , $b=5$ .

…

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CBSE 2026Set ANNUAL1 mark
$i^9 + i^{19}$
›Reveal solutionSolution

Using $i^4=1$ , reduce each exponent mod 4: $i^9=i$ and $i^{19}=-i$ , which sum to 0.

Since $i^4=1$ , any power $i^k = i^{(k \bmod 4)}$ .

$i^9 = i^{(4\times2+1)} = (i^4)^2\cdot i^1 = 1\cdot i = i$ .

…

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CBSE 2026Set ANNUAL1 mark
$5 - 3i$
›Reveal solutionSolution

Modulus of $a+ib$ is $\sqrt{a^2+b^2}$ ; here that gives $\sqrt{34}$ .

For $z=5-3i$ , $a=5$ , $b=-3$ .

…

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CBSE 2026Set 1A1 markMCQ
$\sqrt{-25} \times \sqrt{-9}$
(1) $15$
(2) $-15$
(3) $15i$
(4) None of these
›Reveal solutionSolution

Write $\sqrt{-25}=5i$ and $\sqrt{-9}=3i$ ; their product is $15i^2=-15$ .

The rule $\sqrt{a}\,\sqrt{b}=\sqrt{ab}$ fails for negative numbers, so convert first:

$\sqrt{-25}=5i,\qquad \sqrt{-9}=3i.$ …

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Q.The value of −25×−9\sqrt{-25}\times\sqrt{-9}−25​×−9​ is _____.

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Q.The value of $\sqrt{-25}\times\sqrt{-9}$ is _____.

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