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.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$\sqrt{5}+3i$
(a) $\dfrac{\sqrt{5}}{14}-\dfrac{3}{14}i$
(b) $\dfrac{\sqrt{5}}{14}+\dfrac{3}{14}i$
(c) $\dfrac{3}{14}-\dfrac{\sqrt{5}}{14}i$
(d) $\dfrac{3}{14}+\dfrac{\sqrt{5}}{14}i$
›Reveal solutionSolution

The multiplicative inverse of √5+3i is (√5−3i)/14.

For a complex number z = a+bi, its multiplicative inverse is $\dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-bi}{a^2+b^2}$ .

Here a=√5, b=3, so $a^2+b^2 = 5+9=14$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$i^{-35}$
›Reveal solutionSolution

$i^{-35} = i$ , i.e., in the form a+ib this is 0+1i.

Powers of i cycle with period 4: $i^1=i, i^2=-1, i^3=-i, i^4=1$ , and this pattern repeats.

$i^{-35} = \dfrac{1}{i^{35}}$ . Since $35 = 4\times8+3$ , $i^{35}=i^{3}=-i$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 markMCQ
$z$
(a) $\dfrac{\bar{z}}{|z|^2}$
(b) $\dfrac{\bar{z}}{|z|}$
(c) $\dfrac{z}{|\bar{z}|}$
(d) $\dfrac{z}{|\bar{z}|^2}$
›Reveal solutionSolution

The multiplicative inverse of a nonzero complex number $z$ is $\dfrac{\bar z}{|z|^2}$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 markMCQ
$1 + i + i^2 + i^3 + i^4$
(a) $i$
(b) $0$
(c) $-i$
(d) $1$
›Reveal solutionSolution

$1+i+i^2+i^3+i^4 = 1$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$\left(\dfrac{1}{3} + 3i\right)^3$
›Reveal solutionSolution

$\left(\dfrac13+3i\right)^3 = -\dfrac{242}{27}-26i$ .

Use $(a+bi)^3 = (a^3-3ab^2) + i(3a^2b - b^3)$ with $a=\dfrac13$ , $b=3$ .

Real part: $a^3-3ab^2 = \left(\dfrac13\right)^3 - 3\cdot\dfrac13\cdot3^2 = \dfrac{1}{27} - 9 = \dfrac{1}{27}-\dfrac{243}{27} = -\dfrac{242}{27}$ .

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 markMCQ
$\sqrt{i} + \sqrt{-i}$
(a) 0
(b) 1
(c) 2
(d) $\sqrt{2}$
›Reveal solutionSolution

Writing $i$ and $-i$ in polar form and taking their principal square roots, the imaginary parts cancel on addition, leaving $\sqrt{2}$ .

Write $i$ in polar (trigonometric) form: $i = \cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}$ .

Step 1: Square root of $i$ .

By De Moivre's theorem, the principal square root is

$\sqrt{i} = \cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4} = \dfrac{\sqrt2}{2} + i\dfrac{\sqrt2}{2}$

Step 2: Square root of $-i$ .

Write $-i = \cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)$ , so its principal square root is …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL1 mark
$(i^9 + i^{19})$
›Reveal solutionSolution

Reducing exponents mod $4$ gives $i^9=i$ and $i^{19}=-i$ , which cancel to give $0$ .

Since $i^4=1$ , powers of $i$ repeat every $4$ steps. Reduce each exponent modulo $4$ :

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

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 markMCQ
$\sqrt{i} + \sqrt{-i}$
(a) 0
(b) 1
(c) 2
(d) $\sqrt{2}$
›Reveal solutionSolution

Using the principal square roots of $i$ and $-i$ in polar form, the sum simplifies to $\sqrt{2}$ .

$i$ has modulus $1$ and argument $\dfrac{\pi}{2}$ , so its principal square root is

$\sqrt{i}=\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}(1+i).$

$-i$ has modulus $1$ and argument $-\dfrac{\pi}{2}$ , so its principal square root is …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$3 - 4i$
›Reveal solutionSolution

Rationalising $\dfrac{1}{3-4i}$ using the conjugate $3+4i$ gives $\dfrac{3}{25}+\dfrac{4}{25}i$ .

The multiplicative inverse of $3-4i$ is $\dfrac{1}{3-4i}$ . Multiply numerator and denominator by the conjugate $3+4i$ :

$\dfrac{1}{3-4i}\times\dfrac{3+4i}{3+4i}=\dfrac{3+4i}{3^2+4^2}=\dfrac{3+4i}{9+16}=\dfrac{3+4i}{25}.$ …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$\dfrac{1}{1-i}$
›Reveal solutionSolution

Since $|1-i|=\sqrt2$ , the modulus of its reciprocal is $\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$ .

For any nonzero complex numbers, $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$ .

Here $z_1=1$ (modulus $1$ ) and $z_2=1-i$ , whose modulus is

$|1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}.$

So …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL1 mark
$\sqrt{5} + 3i$
›Reveal solutionSolution

Multiply and divide by the conjugate to get the multiplicative inverse $\dfrac{\sqrt5-3i}{14}$ .

For a complex number $z = a+ib$ , its multiplicative inverse is

$z^{-1} = \dfrac{1}{z} = \dfrac{\bar z}{|z|^2} = \dfrac{a-ib}{a^2+b^2}$

Here $z = \sqrt5 + 3i$ , so $a=\sqrt5,\ b=3$ .

…

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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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