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Question

Question: How do you simplify \[{{i}^{5}}?\]...

How do you simplify i5?{{i}^{5}}?

Explanation

Solution

A complex number is the combination of an imaginary number and the real number.
The given term iiin the question is the complex number and known. A number in the form of a+iba+ibis called a complex number. The term with ii(iota) is called the imaginary part and the first term aais a real part. The complex number is represented by ZZ. And it is expressed as Z=a+ibZ=a+ib.
The value of iiis equal to 1\sqrt{-1}
If the imaginary part of a complex number will be zero then the complex number will be a purely real number. And if the real part of the complex number is zero then the complex number will be purely imaginary.
As we know that the value of iiis equal to 1\sqrt{-1}
i=1\Rightarrow i=\sqrt{-1}.

Complete step by step solution:
The given term in the question is i5{{i}^{5}}
To simplify the above term first of all we must have to know the property of iithat are
We know that i=1i=\sqrt{-1}
i2=i×i=1×1=(1)\Rightarrow {{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}=(-1)
i3=i2×i=(1)×1=1=i\Rightarrow {{i}^{3}}={{i}^{2}}\times i=(-1)\times \sqrt{-1}=-\sqrt{-1}=-i
i4=i2×i2=(1)×(1)=1\Rightarrow {{i}^{4}}={{i}^{2}}\times {{i}^{2}}=(-1)\times (-1)=1
Thus to simplify i5{{i}^{5}}first we have to do the factor 55.
We can write 55in the factors form as
5=2+2+1\Rightarrow 5=2+2+1 or 5=4+15=4+1
Therefore we can rewrite i5{{i}^{5}}, as
i5=i(2+2+1)\Rightarrow {{i}^{5}}={{i}^{(2+2+1)}}
i5=i2×i2×i\Rightarrow {{i}^{5}}={{i}^{2}}\times {{i}^{2}}\times i
i5=(1)×(1)×1\Rightarrow {{i}^{5}}=(-1)\times (-1)\times \sqrt{-1}
i5=1\Rightarrow {{i}^{5}}=\sqrt{-1}
i5=i\Rightarrow {{i}^{5}}=i

Hence the simplified value of i5{{i}^{5}}is ii.

Note:
If we add two complex numbers then each part will be added separately means the imaginary part will be added with the imaginary and the real part will be added with the real part.
For example: Add the following complex number z1=1+6i{{z}_{1}}=1+6iand z2=2+0i{{z}_{2}}=-2+0i
The solution will be z1+z2=1+6i+(2+0i){{z}_{1}}+{{z}_{2}}=1+6i+(-2+0i)
1+(2)+(6+0)i\Rightarrow 1+(-2)+(6+0)i
1+6i\Rightarrow -1+6i.