For a positive integer $n$, find the value of $(1-i)^n\left(1-\dfrac{1}{i}\right)^n$.

For a positive integer n, find the value of (1-i)ⁿ(1-(1)/(i))ⁿ.

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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.

Since $\frac{1}{i} = -i$ , the second factor is $1+i$ , and $(1-i)(1+i)=2$ , so the value is $2^n$ .

Simplify $\dfrac{1}{i}$ :

\frac{1}{i} = \frac{-i}{i\cdot(-i)} = \frac{-i}{1} = -i,

1 - \frac{1}{i} = 1-(-i) = 1+i.

Because both factors share the exponent $n$ :

(1-i)^n\left(1-\frac{1}{i}\right)^n = (1-i)^n(1+i)^n = \big[(1-i)(1+i)\big]^n.

Now

(1-i)(1+i) = 1 - i^2 = 1-(-1) = 2,

hence

(1-i)^n\left(1-\frac{1}{i}\right)^n = 2^n.

✓Final answer

2^n

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Q.For a positive integer nnn, find the value of (1−i)n(1−1i)n(1-i)^n\left(1-\dfrac{1}{i}\right)^n(1−i)n(1−i1​)n.

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Q.For a positive integer $n$ , find the value of $(1-i)^n\left(1-\dfrac{1}{i}\right)^n$ .

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