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Question: Solve the following: \[{i^{57}} + \dfrac{1}{{{i^{125}}}} = ? \] A) 0 B) \[2i\] C) \[ - 2i\] ...

Solve the following: i57+1i125=?{i^{57}} + \dfrac{1}{{{i^{125}}}} = ?
A) 0
B) 2i2i
C) 2i - 2i
D) 2

Explanation

Solution

Here we will find the value of the equation given by using the property of iota which is represented by ii. First, we will use the property that i4=1{i^4} = 1 so for that, we will make the iota term in the power of 4n4n. Then we will solve the iota term by using the exponent property where the exponent with the same base when multiplied there power are added. Finally, we will solve the equation to get the required answer.

Complete step by step solution:
The equation we have to solve is: i57+1i125{i^{57}} + \dfrac{1}{{{i^{125}}}}
As we know
i=1i2=1i3=1i4=1\begin{array}{l}i = \sqrt 1 \\\\{i^2} = - 1\\\\{i^3} = - 1\\\\{i^4} = 1\end{array}
So we can say,
i4n+1=ii4n+2=1i4n+3=ii4n+4=1=i4n\begin{array}{l}{i^{4n + 1}} = i\\\\{i^{4n + 2}} = - 1\\\\{i^{4n + 3}} = - i\\\\{i^{4n + 4}} = 1 = {i^{4n}}\end{array}
Now we will split the term of iota by using exponent property. Therefore, we get
i57+1i125=i4×14i1+1i4×31i1{i^{57}} + \dfrac{1}{{{i^{125}}}} = {i^{4 \times 14}} \cdot {i^1} + \dfrac{1}{{{i^{4 \times 31}} \cdot {i^1}}}
Using i4n=1{i^{4n}} = 1 in the above equation, we get
i57+1i125=i+1i\Rightarrow {i^{57}} + \dfrac{1}{{{i^{125}}}} = i + \dfrac{1}{i}
We will multiply and divide the above equation by ii in the fraction term. So, we get
i57+1i125=i+1i×ii i57+1i125=i+ii2\begin{array}{l} \Rightarrow {i^{57}} + \dfrac{1}{{{i^{125}}}} = i + \dfrac{1}{i} \times \dfrac{i}{i}\\\ \Rightarrow {i^{57}} + \dfrac{1}{{{i^{125}}}} = i + \dfrac{i}{{{i^2}}}\end{array}
As i2=1{i^2} = - 1 we can write above term as,
i57+1i125=i+(i)\Rightarrow {i^{57}} + \dfrac{1}{{{i^{125}}}} = i + \left( { - i} \right)
Adding and subtracting the terms, we get
i57+1i125=0\Rightarrow {i^{57}} + \dfrac{1}{{{i^{125}}}} = 0

Hence, option (A) is correct.

Note:
The numbers which don’t fall anywhere in the number line are known as imaginary numbers also known as complex numbers. The square root of negative numbers is termed Complex number where we denote them in the form of iota. All the numbers having iota i.e. ii in them are known as imaginary numbers. Imaginary numbers is an important concept of mathematics which extends the real number system to the complex number system. We can find the cube root and square root of the complex numbers. Multiplication and division can also be done on complex numbers.