×

Sorry!, This page is not available for now to bookmark.

The value i or the concept of i is used in explaining and expressing complex numbers. Complex numbers are numbers with a real and imaginary part. The imaginary part is defined with the help of i. Basically, “i” is the imaginary part which is also called iota.

Value of i is √-1 A negative value inside a square root signifies an imaginary value. All the basic arithmetic operators are applicable to imaginary numbers. On squaring an imaginary number, we obtain a negative value.

In the study of complex numbers, “i” holds significant importance. We will try to understand the complexities revolving around the value of i with the help of some examples. We will also discuss higher degrees of “i” in detail. With a better understanding of these concepts, students can solve the complex number problems with more ease.

The imaginary part of a complex number is defined as ‘iota’. To calculate the value of an imaginary number, we use the notation iota or “i”. The square root of a negative number gives us an imaginary number.

Value of i = √-1

We use this value of i in understanding the concepts of complex numbers.

For a quadratic equation, x^{2} + 1 = 0

x^{2 }= -1

x = √-1

Here √-1 is the imaginary part.

i = √-1

i^{2 }= -1

As discussed earlier, the squaring of an imaginary number gives us a negative value.

i = √-1

Squaring on both sides

i^{2} = -1

Any complex number is the combination of a real number and an imaginary number. All non-real values are represented by iota or “i”.

Examples of real numbers are 10, -20, √3 etc.

Examples of imaginary numbers are 2i, √-5, -i etc.

Complex numbers are of the form a + ib, where i signifies the imaginary part. Zero is also a complex number. In a complex number, only the real part can be added or subtracted to the real part and only the imaginary part can be added or subtracted with the imaginary part.

Now we will try to understand the value of i in a complex number. For any complex number,

a + ib, a and b are real numbers, where i signifies the imaginary part. Value of i squared

i^{2 }= -1, gives us a negative number.

On multiplying a negative integer to this value we get,

-i^{2 }= 1

If we multiply i to i^{2},we get

i x i^{2 }= i(-1)= -i

On further multiplication, we will get

-i x i = -i^{2 }= -(-1)=1

Given below is a table with the commonly used values of i. Students can refer to these values while solving complex numbers problems.

Memorising all these values can be confusing and tiring! There’s a trick to solving higher degrees of i. The values of i follow a circular pattern. Let’s look at that pattern:

i^{4n }= 1

i^{4n+1 }= i

i^{4n+2} = -1

i^{4n+3 }= -i

Using this circle of formulas, students can calculate the values of i.

For example, at n = 0

i^{0} = 1

i^{1 }= i

i^{2 }= -1

i^{3 }= -i

At n = 1,

i^{5 }= 1

i^{6 }= i

i^{7 }= -1

i^{8} = -i

At n = 2,

i^{9} = 1

i^{10} = i

i^{11 }= -1

i^{12} = -i

The value of higher degrees of i follows this circular formula. Mathematics is a conceptual and application-based study. It doesn’t promote the learning of values. Getting in tune with the concepts will help you understand mathematical applications better.

Complex numbers is a fundamental topic of study. The concept of imaginary numbers is a significant part of it. Students can clear their concepts further by going through these solved examples.

Find the value of 1 + √-3.

Ans: 1 + √-3 is a complex number with a real and imaginary part.

We know that,

√-1 = i, Substituting this value, we get

1 + 3i

It is the simplified form of the equation.

FAQ (Frequently Asked Questions)

Q1. Define the Value of i Power i.

Ans: “i” is an imaginary number, but an imaginary number raised to the power of an imaginary number turns out to be a real number.

The value of i is √-1.

The value of ii is a real number with its value 0.207 approximately. The value of i to the power i is

ii ≃ 0.20788.

Let’s calculate this value mathematically. To calculate the value of i, we will need to understand Euler’s formula first.

According to Euler’s formula,

e^{ix }= cosx + isinx

At x = π/2x = π/2

cis x = cos π/2 + i sin π/2

Here, cis x is just a notation.

Cis x = i

So,

ii = (cis π/2)i

ii = e^{i2 π/2}

ii = e^{-π/2}

ii=0.20788 approximately.

Q2. Solve the following quadratic equation x^{2} + 25 = 0.

Ans: To solve the following quadratic equation, we will have to use the concepts revolving around complex numbers. Here, we will use the treatment of imaginary values to obtain the solution of the equation.

x^{2} + 25 = 0

x^{2}= -25

x = √-25

x = i√25

As the value of i = √-1.

So,

x=i(5)

Therefore, x = 5i will satisfy the equation.

Let’s put this value of x in the equation

x^{2} + 25 = 0

(5i)^{2 }+ 25 = 0

25i^{2 }+ 25 = 0

The value of i^{2} = -1.On substituting this value in the equation, we get

25(-1) + 25 = 0

-25 + 25 = 0

0 = 0

L.H.S = R.H.S, hence proved.