Powers of Iota

What is Iota in Mathematics?

In mathematics, iota ($i$) is a special symbol used to represent the square root of $-1$. Since there is no real number whose square equals $-1$, mathematicians introduced $i$ as an imaginary unit. The powers of iota in mathematics are extremely important because they allow us to work with complex numbers, which are numbers written in the form $a + bi$. Without iota, solving certain quadratic equations or dealing with negative square roots would not be possible. Understanding iota is the first step toward mastering complex numbers.

Value of Iota ($i$)

The square of any real number, whether positive, negative, or zero, is always non-negative, i.e. $x^2 \geq 0$ for all $x \in \mathbb{R}$.

Hence, the equation $x^2 + x + 1 = 0$ is not satisfied for any real value of $x$ and cannot be solved in the real number system. This leads to the concept of imaginary numbers.

Euler was the first mathematician to introduce the symbol $i$ (read as iota). The imaginary number $i$ is defined as the square root of $-1$.

Basic Definition of Iota

From the equation:

$x^2 + 1 = 0 \quad \Rightarrow \quad x^2 = -1$

we get:

$x = \pm \sqrt{-1} = \pm i$

So, the equation $x^2 + 1 = 0$ has two solutions: $x = i$ and $x = -i$.

By definition: $\sqrt{-1} = i \quad \text{and} \quad ( \sqrt{-1} )^2 = i^2 = -1$

Square Root of Negative Numbers Using Iota

The square root of any negative number can be expressed as a multiple of $i$.

For example:

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

Properties and Rules of Iota Powers

The properties and rules of iota powers explain how different exponents of $i$ behave in a cyclic manner. Using these rules, students can simplify complex expressions and quickly determine values of $i^n$ without long calculations. These properties form the foundation for solving higher-level questions in complex numbers.

Fundamental Values of $i$ ($i^1, i^2, i^3, i^4$)

The four fundamental powers of iota are the backbone of all calculations:

• $i^1 = i$

• $i^2 = -1$

• $i^3 = -i$

• $i^4 = 1$

Once we reach $i^4 = 1$, the cycle begins again. So, $i^5 = i$, $i^6 = -1$, and so on. These four values are the building blocks of all powers of iota. Many exam questions directly test these fundamental results, so memorizing them is crucial.

Value of Powers of Iota

The value of powers of iota helps in understanding how $i^n$ behaves for different exponents. By knowing the standard results of $i^0, i^1, i^2, i^3,$ and $i^4$, students can easily find the value of any higher power of $i$ using its cyclic pattern.

Case 1: Positive Integer Powers of $i$

The powers of $i$ follow a repeating cycle of four:

$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \dots$

In general:

$i^{4n} = 1, \quad i^{4n+1} = i, \quad i^{4n+2} = -1, \quad i^{4n+3} = -i$

Case 2: Negative Integer Powers of $i$

Negative powers can be simplified using reciprocals:

$i^{-1} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$

$i^{-2} = \frac{1}{i^2} = -1$

$i^{-3} = \frac{1}{i^3} = \frac{1}{-i} = i$

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

Thus, negative powers also follow the same repeating cycle.

Cyclic Nature of Iota Powers ($i^n$ repeats every 4)

One of the most interesting properties of iota is that its powers are cyclic with a period of 4. This means after every 4 steps, the sequence of values repeats itself. For example:

• $i^1 = i$

• $i^2 = -1$

• $i^3 = -i$

• $i^4 = 1$

• $i^5 = i$ (same as $i^1$)

• $i^6 = -1$ (same as $i^2$)

This cycle continues infinitely. Because of this repetition, no matter how large the power of iota is, we only need to check its remainder when divided by 4.

Standard Formula for Calculating $i^n$

To calculate higher powers of iota quickly, use the remainder rule:

• If $n \mod 4 = 1$, then $i^n = i$

• If $n \mod 4 = 2$, then $i^n = -1$

• If $n \mod 4 = 3$, then $i^n = -i$

• If $n \mod 4 = 0$, then $i^n = 1$

For example:

• $i^{27}$ → remainder of $27 \div 4$ is $3$, so $i^{27} = -i$

• $i^{56}$ → remainder of $56 \div 4$ is $0$, so $i^{56} = 1$

This formula makes solving large powers very efficient, and that is why it is often tested in exams.

Special Property: Sum of Four Consecutive Powers

For any integer $n$:

$i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n (1 + i + i^2 + i^3)$

Since $(1 + i + i^2 + i^3) = (1 + i - 1 - i) = 0$:

$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$

This property shows that the sum of any four consecutive powers of $i$ is always zero.

Powers of Iota with Values

The powers of iota with values show how $i^1, i^2, i^3, i^4$ follow a repeating cycle. Learning these values makes it easier to simplify higher powers of $i$ in complex number problems and quickly solve exam-based questions.

Step-by-Step Calculation of Iota Powers

We can build the powers of iota table step by step:

• $i^1 = i$

• $i^2 = -1$

• $i^3 = -i$

• $i^4 = 1$

• $i^5 = i$

• $i^6 = -1$

• $i^7 = -i$

• $i^8 = 1$

And so on. This table helps in visualizing the repetition of powers and is useful for quick reference during practice.

Tricks to Remember Iota Powers Easily

• Always remember the repeating cycle: $i, -1, -i, 1$

• Think of it like moving in a loop of 4 steps

• For larger exponents, just divide by 4 and check the remainder

• Use mental shortcuts: odd powers give $i$ or $-i$, even powers give $-1$ or $1$

These tricks make powers of iota one of the easiest topics to score in exams.

Key Observations from the Iota Cycle

• The cycle of iota is fixed and never changes

• Powers of $i$ can only take 4 possible values: $i, -1, -i, 1$

• Higher powers can always be reduced to one of these 4 values

• This property is the reason why iota is simple to work with despite being imaginary.

Solved Examples Based on Powers of Iota

Example 1: If $a<0$ and $b<0$, then $\sqrt{a }\sqrt{ b}$ is equal to:

Solution:

Here, $a$ and $b$ are both negative

$
\begin{aligned}
& \sqrt{-a} \sqrt{-b}=\sqrt{(-1) a} \sqrt{(-1) b} \\
& \Rightarrow \sqrt{-1} \sqrt{a} \sqrt{-1} \sqrt{b}=i^2 \sqrt{a} \sqrt{b} \\
& \Rightarrow-1 \sqrt{|a||b|}
\end{aligned}
$
so the value is $-\sqrt{(|a||b|)}$
Hence, the answer is $-\sqrt{(|a||b|) }$.

Example 2: $(1+i)^8+(1-i)^9$ equals:

Solution:

$
(1+i)^8=\left((1+i)^2\right)^4=\left(1+i^2+2 i\right)^4=(1-1+2 i)^4=(2 i)^4=2^4 \cdot i^4=16
$
and
$
\begin{aligned}
& (1-i)^9=\left((1-i)^2\right)^4 \cdot(1-i)^1=\left(1+i^2-2 i\right)^4 \cdot(1-i) \\
& =(-2 i)^4(1-i)=(-2)^4 \cdot i^4 \cdot(1-i)=16(1-i)=16-16 i \\
& \therefore(1+i)^8+(1-i)^9=16+16-16 i=32-16 i
\end{aligned}
$

Hence, the answer is $32-16 i$

Example 3: The value of $i^{-99}$ is:

Solution:

$i^{-99}=\frac{1}{i^{99}}=\frac{i}{i^{100}}=i \quad\left(\because i^{100}=\left(i^4\right)^{25}=1\right)$

So the value is $i$

Hence, the answer is $i$.

Example 4: If $\left(\frac{1+i}{1-i}\right)^x=1$, then $x=$ ?

Solution:

Power of i in Complex Numbers -

$i^{4 n}=1, i^{4 n+1}=i, i^{4 n+2}=-1, i^{4 n+3}=-i$, where $n \in$ Integer
Now,
$
\begin{aligned}
& \left(\frac{1+i}{1-i}\right)^x=\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^x=\left[\frac{(1+i)(1+i)}{2}\right]^x \\
& =\left[\frac{(1+i)^2}{2}\right]^x=\left[\frac{1-1+2 i}{2}\right]^x=(i)^x
\end{aligned}
$

And given that
$
\begin{aligned}
& \left(\frac{1+i}{1-i}\right)^x=1 \\
& i^x=(i)^{4 n} \\
& x=4 n
\end{aligned}
$

Hence, the answer is $4n$.

Example 5: The number of the value of $i^n + i^{-n}$ for different n ∈ I (integer) and $i = \sqrt{-1}$.

Solution:

$ i^{n} + i^{-n} = i^{n} + \frac{1}{i^{n}} $

$ n = 1 $

$\Rightarrow i^1 + \frac{1}{i^1} = i - i = 0 $

$ n = 2 $

$\Rightarrow i^2 + \frac{1}{i^2} = -1 - 1 = -2 $

$ n = 3 $

$\Rightarrow i^3 + \frac{1}{i^3} = -i + i = 0 $

$ n = 4 $

$\Rightarrow i^4 + \frac{1}{i^4} = 1 + 1 = 2 $

$ n = 5 $

$\Rightarrow i^5 + \frac{1}{i^5} = i - i = 0 $

So the values are $0,-2$, and 2.

List of topics related to Powers of Iota

Below is a list of important topics related to powers of iota in mathematics, including definitions, rules, properties, and solved examples. These topics help students quickly understand and remember how to solve problems involving powers of $i$ in complex numbers.

NCERT Resources

Below are the NCERT resources for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations, which include NCERT solutions, exemplar problems, and detailed notes. These resources help students strengthen their understanding of powers of iota, properties of complex numbers, and quadratic equations with step-by-step explanations.

NCERT Solutions for Class 11 Maths Chapter 5 -Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Class 11 Maths Notes for Chapter 5 - Complex Numbers and Quadratic Equations

Practice Questions based on Powers of Iota

Below are some practice questions based on powers of iota to help students apply formulas and rules effectively. Solving these powers of $i$ problems will improve speed, accuracy, and understanding of the cyclic nature of iota powers in complex numbers.

Powers Of Iota - Practice Question MCQ

We have shared pratice questions sets for related topics below: