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MathematicsClass 11Mathematics

Mathematics | Class 11 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 2 min read

Mathematics – this guide gives you a concise, exam-ready overview of Mathematics from Class 11 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Historical Note

This section provides a historical perspective on the development of complex numbers. The Greeks recognized that square roots of negative numbers do not exist within the real number system. The Indian mathematician Mahayira, around 850 AD, first clearly stated this difficulty in his work 'Ganitasara Sangraha', noting that negative quantities are not squares and thus have no square roots. Bhaskara, another Indian mathematician, reiterated this in his work 'Bijaganita' around 1150 AD.

In 1545, Cardan studied equations like x + y = 10 and xy = 40 and obtained solutions involving square roots of negative numbers, which he dismissed as 'useless'. Later, Albert Girard in the early 17th century accepted square roots of negative numbers, recognizing their importance in obtaining all roots of polynomial equations.

Euler introduced the symbol i for √-1, and W. R. Hamilton, around 1830, gave a rigorous mathematical definition of complex numbers as ordered pairs of real numbers, avoiding the term 'imaginary numbers'. This historical development highlights the gradual acceptance and formalization of complex numbers in mathematics.

📊 Diagram: Image of W. R. Hamilton (1805-1865) referenced earlier.

🧪 Activity: No activity in this section.

🔗 Connection: Provides historical context for the entire chapter on complex numbers.

Frequently asked questions

1. $(5i)\left(-\frac{3}{5}i\right)$

Given: (5i)\left(-\frac{3}{5}i\right)

Step 1: Multiply the constants: 5 (-3/5) = -3 Step 2: Multiply the imaginary units: i i = i^2 = -1 Step 3: So, (5i)(-3/5 i) = -3 * (-1) = 3

Answer: 3

2. $i^9 + i^{19}$

Recall powers of i cycle every 4: i^1 = i i^2 = -1 i^3 = -i i^4 = 1

Calculate i^9: 9 mod 4 = 1, so i^9 = i

Calculate i^{19}: 19 mod 4 = 3, so i^{19} = -i

Sum: i + (-i) = 0

Answer: 0

3. $i^{-39}$

Recall powers of i cycle every 4. First, convert negative power: i^{-39} = 1 / i^{39}

Calculate 39 mod 4: 39 mod 4 = 3 So, i^{39} = i^3 = -i

Therefore, i^{-39} = 1 / (-i) = -1 / i

Multiply numerator and denominator by i: = (-1 / i) * (i / i) = (-i) / (i^2) = (-i) / (-1) = i

Answer: i

4. $3(7 + i7) + i(7 + i7)$

Given expression: 3(7 + 7i) + i(7 + 7i)

Step 1: Expand each term: 3(7 + 7i) = 21 + 21i

Step 2: i(7 + 7i) = 7i + 7i^2 = 7i + 7(-1) = 7i - 7

Step 3: Add the two results: (21 + 21i) + (7i - 7) = (21 - 7) + (21i + 7i) = 14 + 28i

Answer: 14 + 28i

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