# Powers of iota

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 1 What is iota and Its value? 2 Square Root of iota 3 Important Notes on Powers of iota 4 Higher Powers of iota 5 Tips and Tricks on Powers of iota 6 Powers of iota Calculator 7 Solved Examples on Powers of iota 8 Practice Question 9 Maths Olympiad Sample Papers 10 Frequently Asked Questions (FAQs)

## What is Iota and Its Value?

Iota is a complex number that is denoted by $$\mathbf{i}$$ and the value of iota is $$\mathbf{\sqrt{-1}}$$.

i.e.,

 $i= \sqrt{-1}$

If we square both sides of the above equation, we get:

$i^2=-1$

i.e., the value of the square of iota is -1

While solving quadratic equations, you might have come across situations where the discriminant is negative.

For example, consider the quadratic equation

$x^2+x+1=0$

If we use the quadratic formula to solve this, we get the discriminant (the part inside the square root) as a negative value.

In such cases, we write $$\sqrt{-3}$$ as:

$\sqrt{-3} =\sqrt{-1} \times \sqrt{3}= i \times \sqrt{3}= \sqrt{3}i$

This would give the solution of the above quadratic equation to be:

$x=\dfrac{-1 \pm \sqrt{3}i}{2}$

Hence, the value of iota is helpful in solving square roots with negative values.

Thus,

 The value of iota is, $$i= \sqrt{-1}$$ The square of iota is, $$i^2=-1$$

## Square Root of Iota

We will find the value of the square root of iota, i.e., $$\sqrt{i}$$

We will use De Moivre's Theorem to find this.

We know that,

\begin{align}i&= \cos \dfrac{\pi}{2}+ i \sin \dfrac{\pi}{2}\0.3cm] &= \cos \left( \dfrac{\pi}{2}+2n\pi \right)+ i \sin \left(\dfrac{\pi}{2}+2n\pi\right),n=0,1 \\[0.3cm] &= \cos \left( \dfrac{\pi+4n\pi}{2} \right)+ i \sin \left( \dfrac{\pi+4n\pi}{2} \right) \end{align} Here, we took $$n=0,1$$ as we need 2 solutions. But we have to find $$\sqrt{i}=(i)^{1/2}$$ Let us raise the exponent to $$1/2$$ on both sides. So we get: \begin{align} \sqrt{i}&=\left(\cos \left( \dfrac{\pi+4n\pi}{2} \right)+ i \sin \left( \dfrac{\pi+4n\pi}{2} \right) \right)^{1/2}\\[0.3cm] &=\cos \left( \dfrac{\pi+4n\pi}{4} \right)+ i \sin \left( \dfrac{\pi+4n\pi}{4} \right),n=0,1 \end{align} When $$n=0$$, $$\sqrt{i}= \cos \dfrac{\pi}{4}+ i \sin \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}+ i \dfrac{\sqrt{2}}{2}$$. When $$n=1$$, $$\sqrt{i}= \cos \dfrac{5\pi}{4}+ i \sin \dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}- i \dfrac{\sqrt{2}}{2}$$  \[ \begin{align} \sqrt{i} &= \dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2} \\[0.2cm] &=-\dfrac{\sqrt{2}}{2}-i \dfrac{\sqrt{2}}{2} \end{align}

Important Notes
1. The value of iota is $$\mathbf{i=\sqrt{-1}}$$
2. The value of the square of iota is, $$\mathbf{i^2=-1}$$
3. The value of the square root of iota is,
\mathbf{ \begin{align} \sqrt{i} &= \dfrac{\sqrt{2}}{2}+i \dfrac{\sqrt{2}}{2} \\[0.2cm] &=-\dfrac{\sqrt{2}}{2}-i \dfrac{\sqrt{2}}{2} \end{align}}

## Higher Powers of Iota

We know that $$i^2=-1$$

Hence, $$-1$$ is the square of iota.

Let us see how to calculate some other powers of iota.

$\begin{array}{l} i^{3}=i \times i^{2}=i \times-1=-i \\ i^{4}=i^{2} \times i^{2}=-1 \times-1=1 \\ i^{5}=i \times i^{4}=i \times 1=i \\ i^{6}=i \times i^{5}=i \times i=i^{2}=-1 \\ i^{7}=i \times i^{6}=i \times-1=-i \\ i^{8}=\left(i^{2}\right)^{4}=(-1)^{4}=1 \\ i^{9}=i \times i^{8}=i \times 1=i \\ i^{10}=i \times i^{9}=i \times i=i^{2}=-1 \end{array}$

But what if the power of iota is a larger number?

If we have to find this value using the previous procedure, it will take quite some time and effort.

There must be some other method to find this value easily.

If we observe all the powers of iota in the above equations, we can derive the following rules:

 $\begin{array}{l}i^{4 k}=1 \\i^{4 k+1}=i \\i^{4 k+2}=-1 \\i^{4 k+3}=-i\end{array}$

Here, $$k$$ is any whole number.

The following figure represents the values for various powers of iota in the form of a continuous circle.

Let us understand how to find the value using an example.

To find $$i^{20296}$$, we first divide 20296 by 4 and find the remainder.

The remainder is 0 (by divisibility rules, we can just divide the number formed by the last two digits which is 96 in this case, to find the remainder).

Thus, using the above rules,

$i^{20296} = i^0 = 1$

$\Rightarrow \boxed{\left( {i^{20296}} \right) = {4}}$

We will just have to remember that $$i^2=-1$$ and $$i^3=-i$$

We will find some other higher powers of iota using these and the above rules.

$$\mathbf{i^n}$$

Remainder

(when $$\mathbf{n}$$ is divided by 4)

Value

$$i^{517}$$

1 $$i^{517} =i^1= \color{blue}{\boxed{\mathbf{i}}}$$

$$i^{2095}$$

3 $$i^{2095}=i^3 = \color{blue}{\boxed{\mathbf{-i}}}$$

$$i^{23456}$$

0 $$i^{23456}=i^0= \color{blue}{\boxed{\mathbf{1}}}$$

$$i^{324770}$$

2 $$i^{324770}=i^2= \color{blue}{\boxed{\mathbf{-1}}}$$

But what if the exponent is negative?

We first convert it into a positive exponent using the negative exponent law and then we apply the rule:

$\dfrac{1}{i}=-i$

This is because

\begin{align}\dfrac{1}{i}&=\dfrac{1}{i}\cdot \dfrac{i}{i}\\[0.2cm]&= \dfrac{i}{i^2}\\[0.2cm]&=\dfrac{i}{-1}\\[0.2cm]&=-i \end{align}

We will use this to find the value of $$i^{-3927}$$

\begin{align}i^{-3927} &=\frac{1}{i^{3927}}\left[\because a^{-m}=\frac{1}{a^{m}}\right] \\[0.2cm]&=\frac{1}{i^{3}} \\&\quad[\because \text {remainder of } 3927 \text { divided by } 4 \text { is } 3]\\[0.2cm]&=\frac{1}{-i} \,\, [\because i^3=-i]\\[0.2cm]&=-\frac{1}{i} \\[0.2cm]&=-(-i) \,\, [\because \frac{1}{i}=-i] \\[0.2cm]&=i\end{align}

Tips and Tricks
1. To find any power of iota, say $$\mathbf{i^n}$$, just divide $$\mathbf{n}$$ by 4 and find the remainder, r
Then just apply $$\mathbf{i^n=i^r}$$
Here, you just need to remember two things $$\mathbf{i^2=-1}$$ and $$\mathbf{i^3=-i}$$
2. To calculate the negative powers of iota, we use the rule $$\mathbf{\dfrac{1}{i}=-i}$$

## Powers of Iota Calculator

Here is the "Powers of Iota Calculator".

Here, you can enter any exponent (positive or negative) of $$i$$ and see the result in a step-by-step manner.

## Solved Examples

 Example 1

Find the values of the following:

$i^{37}, \,\,i^{99}, \,\,i^{-1}, \,\,i^{-50}$

Solution:

Since $${i^2} = - 1$$, we have:

\begin{align}&{i^3} = {i^2} \times i = - 1 \times i = - i\\&{i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1\end{align}

Now, we can calculate i raised to any integer power. For example,

\begin{align}&{i^{37}} = {i^{36}} \times i = {\left( {{i^4}} \right)^9} \times i = 1 \times i = i\\&{i^{99}} = {i^{96}} \times {i^3} = {\left( {{i^4}} \right)^{24}} \times {i^3} = 1 \times - i = - i\\&{i^{ - 1}} = \frac{1}{i} = \frac{i}{{{i^2}}} = \frac{i}{{ - 1}} = - i\\&{i^{ - 50}} = \frac{1}{{{i^{50}}}} = \frac{1}{{{i^{48}} \times {i^2}}} = \frac{1}{{{i^2}}} = - 1\end{align}

Thus,

 \begin{align}{i^{37}} &= i\\{i^{99}}& = - i\\{i^{ - 1}}&= - i\\{i^{ - 50}} &= - 1\end{align}
 Example 2

Find the value of $${i^{4n + k}}$$, where n and k are integers, and k is in the set {0, 1, 2, 3}

Solution:

We have:

${i^{4n + k}} = {i^{4n}} \times {i^k} = {\left( {{i^4}} \right)^n} \times {i^k} = 1 \times {i^k} = {i^k}$

Thus, the value of $${i^{4n + k}}$$ is the same as the value of $${i^k}$$, which depends on the value of k:

\begin{align}&k = 0:\; {i^k} = \;1\\&k = 1:\; {i^k} = \;i\\&k = 2:\; {i^k} = - 1\\&k = 3: \;{i^k} = - i\end{align}

Thus,

 $\begin{array}{l} k=0: i^{4n+k}=1 \\ k=1: i^{4n+k}=i \\ k=2: i^{4n+k}=-1 \\ k=3: i^{4n+k}=-i \end{array}$
 Example 3

(a) Find the value of \begin{align}{i^{500}} + {i^{501}} + {i^{502}} + {i^{503}}\end{align}

(b) Show that the sum of any four consecutive powers of iota is 0

Solution:

(a) We have:

\begin{align}&{i^{500}} = {\left( {{i^4}} \right)^{125}} = {1^{125}} = 1\\&{i^{501}} = {i^{500}} \times i = i\\&{i^{502}} = {i^{500}} \times {i^2} = {i^2} = - 1\\&{i^{503}} = {i^{500}} \times {i^3} = {i^3} = - i\end{align}

Clearly, the sum of these four terms is 0

(b) We have:

\begin{align}&{i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}\\&= {i^n}\left( {1 + i + {i^2} + {i^3}} \right)\\&= {i^n}\left( {1 + i - 1 - i} \right) = 0\end{align}

Thus,

## 2. What is iota cube?

$$i^3= i^2 \cdot i = (-1) \cdot i= -i$$

Thus, iota cube is $$-i$$

## 3. What is the value of i cube?

$$i^3= i^2 \cdot i = (-1) \cdot i= -i$$

Thus, $$i$$ cube is $$-i$$

Complex Numbers
grade 10 | Questions Set 1
Complex Numbers
Complex Numbers