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Question

Question: How do you simplify \({i^{18}}\)?...

How do you simplify i18{i^{18}}?

Explanation

Solution

Let us first know what a complex number is. Complex numbers are any number in the form of Z=a+ibZ = a + ib where aa and bb are real numbers and ii is an imaginary number known as iota. It is equal to 1\sqrt { - 1} . Complex numbers are denoted as ZZ.
It must be known that when ii is raised to certain powers, it assumes some values which are given as:
i=1i = \sqrt { - 1}
i2=1{i^2} = - 1 (As i2=(1)2=1{i^2} = {(\sqrt { - 1} )^2} = - 1)
i3=i{i^3} = - i (As i3=i2×i=1×i=i{i^3} = {i^2} \times i = - 1 \times i = - i)
i4=1{i^4} = 1 (As i4=i2×i2=1×1=1{i^4} = {i^2} \times {i^2} = - 1 \times - 1 = 1)
We will use these 44 values of ii to solve the given question.

Complete step by step solution:
Given expression is i18{i^{18}}. We will now simplify it by breaking the exponent 1818into factors, such that
i18=i2×3×3\Rightarrow {i^{18}} = {i^{2 \times 3 \times 3}}
We will now separate the factors of the exponent into 22 and all other factors, such that
i18=i2×9\Rightarrow {i^{18}} = {i^{2 \times 9}}
We will now use one of the law of exponents which states xp×q=(xp)q{x^{p \times q}} = {({x^p})^q}, such that
i18=(i2)9\Rightarrow {i^{18}} = {({i^2})^9}
On substituting the value of i2=1{i^2} = - 1, we will get
i18=(1)9\Rightarrow {i^{18}} = {( - 1)^9}
It must be known that if 1 - 1 is raised to the power of an odd number, we will get 1 - 1 itself as the answer. Since 99 is an odd number, therefore we will get
i18=1\Rightarrow {i^{18}} = - 1

Hence, on simplifying i18{i^{18}}, we get 1 - 1 as the answer.

Note:
The given question can also be solved in an alternate way. We know that i4=1{i^4} = 1. So we will simplify the exponent 1818of the expression i18{i^{18}} in the form of 4m+n4m + n where mm and nn are whole numbers. Hence on simplifying, we will get
i18=i4×4+2\Rightarrow {i^{18}} = {i^{4 \times 4 + 2}}
We will now use one of the law of exponents which states xp+q=xp×xq{x^{p + q}} = {x^p} \times {x^q}, such that
i18=i4×4×i2\Rightarrow {i^{18}} = {i^{4 \times 4}} \times {i^2}
Again we will use one of the laws of exponents which states xp×q=(xp)q{x^{p \times q}} = {({x^p})^q}, such that
i18=(i4)4×i2\Rightarrow {i^{18}} = {({i^4})^4} \times {i^2}
We know that i4=1{i^4} = 1. Therefore on substituting this value in the above equation, we will get
i18=(1)4×i2\Rightarrow {i^{18}} = {(1)^4} \times {i^2}
i18=1×i2\Rightarrow {i^{18}} = 1 \times {i^2}
So i18{i^{18}} gets reduced as given,
i18=i2\Rightarrow {i^{18}} = {i^2}
Now we know that i2=1{i^2} = - 1. Therefore on substituting this value in the above equation, we will get
i18=1\Rightarrow {i^{18}} = - 1
Hence on simplifying i18{i^{18}}, we got 1 - 1.