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Question: Evaluate the following \({{i}^{403}}\)...

Evaluate the following i403{{i}^{403}}

Explanation

Solution

Now consider the given number i403{{i}^{403}} we know that the value of i4=1{{i}^{4}}=1 and i3=−i{{i}^{3}}=-i hence we will first simplify the power and write the number in the form of i3{{i}^{3}} and i4{{i}^{4}} by using the laws of indices xm+n=xmxn{{x}^{m+n}}={{x}^{m}}{{x}^{n}} and (xm)n=xmn{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}} . Now we will substitute the values of i3{{i}^{3}} and i4{{i}^{4}} and hence find the solution.

Complete step by step answer:
Now let us first understand the meaning of letter i.
Now we know the number line which represents real numbers which are either rational or irrational.
But there are also numbers which are not real. These numbers are called complex numbers.
Complex numbers are numbers of the form a + ib. where a and b are real and the letter i denotes iota which is nothing but −1\sqrt{-1} .
Now since we have i=−1i=\sqrt{-1} squaring both the sides we get i2=−1{{i}^{2}}=-1
Now multiplying I on both sides we get, i3=−i{{i}^{3}}=-i again multiplying i on both sides we get, i4=−i×i=−(−1)=1{{i}^{4}}=-i\times i=-\left( -1 \right)=1
Hence we can say that i4=1{{i}^{4}}=1
Now consider the given number i403{{i}^{403}} .
Now we know by law of indices that xm+n=xmxn{{x}^{m+n}}={{x}^{m}}{{x}^{n}}
Hence we can write
⇒i403=i400+3=i400i3\Rightarrow {{i}^{403}}={{i}^{400+3}}={{i}^{400}}{{i}^{3}}
Now again by law of indices we know that xmn=(xm)n{{x}^{mn}}={{\left( {{x}^{m}} \right)}^{n}} hence using this we get,
⇒i403=(i4)100i3\Rightarrow {{i}^{403}}={{\left( {{i}^{4}} \right)}^{100}}{{i}^{3}}
Now since i4=1{{i}^{4}}=1 and i3=−1{{i}^{3}}=-1 we will substitute the values in the equation,
⇒i403=1100(−i)=−i\Rightarrow {{i}^{403}}={{1}^{100}}\left( -i \right)=-i
Hence the value of i403{{i}^{403}} is – i.

Note: Note that any power to i will be either of −1,−i,1,i-1,-i,1,i as after i4{{i}^{4}} same values of I will keep repeating. Hence we can find any power of i by just using the laws of indices and the known values of i. Also note that here we also get 1 and -1 as solutions which means the square of a complex number is real. Hence we can say that multiplication of two complex numbers can be real numbers.