i¹⁹⁴⁷ + i¹⁹⁵⁰

Exercise 2.1 · Q1

i^{1947} + i^{1950}

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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

Complex numbers form a field — they obey the same arithmetic rules as real numbers (commutative, associative, distributive) with one extra rule: $i^2 = -1$ . Every operation reduces to real-number arithmetic plus that single rule.

Important

Complex Number Arithmetic — the set of all numbers $a + bi$ with $a, b \in \mathbb{R}$ and $i^2 = -1$ , with addition and multiplication defined as above. This system is closed under all four basic operations (except division by zero), and every non-zero complex number has a multiplicative inverse.

You'll use these operations constantly in everything from solving quadratic equations to analyzing AC circuits to understanding quantum mechanics. Master them now, and the rest becomes much easier.

Complex number arithmetic, including addition, multiplication, and division using the conjugate, is a central skill in the NCERT Class 11 Mathematics chapter on Complex Numbers and Quadratic Equations, and "complex number arithmetic operations with examples" is a heavily searched revision topic for CBSE boards and JEE Main. This arithmetic is foundational for solving polynomial equations with no real roots, a question type that appears often in "complex numbers important questions" for competitive exams.

Powers of $i$ repeat with period $4$ , so every exponent can be replaced by its remainder on division by $4$ ; we do this for both terms and add.

Step 1. Reduce $1947$ mod $4$ . $1947=4\times486+3$ , so the remainder is $3$ , hence $i^{1947}=i^3=-i$ .

Step 2. Reduce $1950$ mod $4$ . $1950=4\times487+2$ , so the remainder is $2$ , hence $i^{1950}=i^2=-1$ .

Step 3. Add the two results.

i^{1947}+i^{1950}=-i+(-1)=-1-i.

✓Final answer

i^{1947}+i^{1950}=\boxed{-1-i}

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 $\sqrt5 - 3i$ :

$\dfrac{1}{\sqrt5+3i} = \dfrac{\sqrt5-3i}{(\sqrt5+3i)(\sqrt5-3i)} = \dfrac{\sqrt5-3i}{(\sqrt5)^2 - (3i)^2} = \dfrac{\sqrt5-3i}{5-(-9)} = \dfrac{\sqrt5-3i}{14}$

$= \dfrac{\sqrt5}{14} - i\cdot\dfrac{3}{14}$

✓Final answer
$\dfrac{\sqrt5}{14} - i\cdot\dfrac{3}{14}$
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.

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

Similarly, $\dfrac{1-i}{1+i}$ is the conjugate expression, giving $-i$ .

So the expression becomes:

$i - (-i) = 2i$

Modulus of $2i$ is $|2i| = \sqrt{0^2+2^2} = 2$ .

✓Final answer
$(a) 2.$
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 $i^m = 1$ .

The powers of $i$ cycle with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ , and then repeat. The smallest positive integer $m$ for which $i^m = 1$ is $m = 4$ .

✓Final answer
$(d) 4.$
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: $i, -1, -i, 1, i, -1, \dots$

Since $28 = 4 \times 7$ is exactly divisible by 4, $i^{28} = (i^4)^7 = 1^7 = 1$ .

✓Final answer
$1$
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.

Here $z = 2-3i$ , so $a=2$ , $b=-3$ . Its conjugate is $\bar z = 2-(-3)i = 2+3i$ .

✓Final answer
$2-3i$
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:

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

Check against the given options: (a) $1-\sqrt3i$ — wrong scale. (b) $\dfrac{1-\sqrt3i}{2}$ — wrong denominator (should be 4, not 2). (c) $\dfrac{1+\sqrt3i}{4}$ — correct denominator but wrong sign (this is $z$ itself divided by 4, not $\bar z$ divided by 4). None of (a), (b), (c) equal the correct value $\dfrac{1-\sqrt3i}{4}$ .

✓Final answer
$\dfrac{1-\sqrt3 i}{4}$
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 $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ :

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

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

(Check by rationalizing: $\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i} = \dfrac{(1+i)^2}{2} = \dfrac{2i}{2}=i$ , and $|i|=1$ ✓.)

✓Final answer
$\dfrac{1+i}{1-i}$
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: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ .

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

$i^{19} = i^{4\times4+3} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i$ .

So $i^9 \cdot i^{19} = i \cdot (-i) = -i^2 = -(-1) = 1$ .

✓Final answer
$i^9 \cdot i^{19} = 1$
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 $a=3$ , $b=5$ .

By definition, $\bar{z} = a - ib = 3-5i$ .

✓Final answer
$3-5i$
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 $i^k = i^{(k \bmod 4)}$ .

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

$i^{19} = i^{(4\times4+3)} = (i^4)^4\cdot i^3 = 1\cdot(-i) = -i$ .

Adding: $i^9+i^{19} = i + (-i) = 0$ .

✓Final answer
$i^9+i^{19}=0$
CBSE 2026Set ANNUAL1 mark
$5 - 3i$
›Reveal solutionSolution

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

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

$|z| = \sqrt{a^2+b^2} = \sqrt{5^2+(-3)^2} = \sqrt{25+9} = \sqrt{34}$ .

✓Final answer
$|5-3i| = \sqrt{34}$
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:

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

Then $\sqrt{-25}\times\sqrt{-9}=5i\times 3i=15i^2=15(-1)=-15.$

✓Final answer
$-15$

←Previous section · 2.1.1Powers of Imaginary Unit i →Next question · 2.1.1Q2 — i¹⁹⁴⁸ - i⁻¹⁸⁶⁹

Q.i1947+i1950i^{1947} + i^{1950}i1947+i1950

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Q. $i^{1947} + i^{1950}$

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