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Question

Question: How do you simplify \[{i^{35}}\]?...

How do you simplify i35{i^{35}}?

Explanation

Solution

Given simple can be solved if we know the laws of indices and the value of ii if it has an even power as power. Numbers are of two types – real numbers and imaginary numbers, the numbers that can be plotted on a number line are called real numbers while the imaginary numbers, as the name suggests, cannot be represented on the number line. As we know that the square root of a negative number doesn’t exist so we had to think of a way to represent them that’s why we take 1=i\sqrt { - 1} = i. We know the law of brackets, that is (xm)n=xm×n{\left( {{x^m}} \right)^n} = {x^{m \times n}}. Using this we can solve this.

Complete step by step answer:
Given i35{i^{35}}.
We can write 35 as 35=32+2+135 = 32 + 2 + 1
i35=i32+2+1{i^{35}} = {i^{32 + 2 + 1}}
Applying law of multiplication, that is xm×xn=xm+n{x^m} \times {x^n} = {x^{m + n}}
i32×i2×i\Rightarrow {i^{32}} \times {i^2} \times i
We can write 32 as 32=2×1632 = 2 \times 16, because we know i2=1{i^2} = - 1.
i2×16×i2×i\Rightarrow{i^{2 \times 16}} \times {i^2} \times i
Applying the law of brackets, that is (xm)n=xm×n{\left( {{x^m}} \right)^n} = {x^{m \times n}}.
(i2)16×i2×i\Rightarrow{\left( {{i^2}} \right)^{16}} \times {i^2} \times i
We know that i2=1{i^2} = - 1, substituting we have,
(1)16×1×i\Rightarrow {\left( { - 1} \right)^{16}} \times - 1 \times i
1×1×i\Rightarrow 1 \times - 1 \times i
i\Rightarrow - i

Thus we have i35=i{i^{35}} = - i.

Note: We also know that odd power of any negative number is always negative and that even power of any negative number is positive number. In mathematics we should know the laws of indices.
Law of multiplication: If the two terms have the same base and are to be multiplied together their indices are added. That is xm×xn=xm+n{x^m} \times {x^n} = {x^{m + n}}.
Law of division: If the two terms have the same base and are to be divided their indices are subtracted. That is xmxn=xmn\dfrac{{{x^m}}}{{{x^n}}} = {x^{m - n}}.
Law of brackets: If a term with a power is itself raised to a power then the powers are multiplied together. That is (xm)n=xm×n{\left( {{x^m}} \right)^n} = {x^{m \times n}}. We use this in most of the mathematical problems, remember them.