Question

Show that \[1+{{i}^{10}}+{{i}^{20}}+{{i}^{30}}\] is a real number.

Answer 1

Hint: As we know that a real number includes all rational and irrational numbers. It also includes positive numbers, negative numbers and the number zero. For the above we will have to know about the properties of iota \[(i)\] given as follows:
\[\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{4n+2}}=-1 \\
 & {{i}^{4n+1}}=i \\
 & {{i}^{4n+3}}=-i \\
\end{align}\]
Here n can be 0, 1,2, 3…

Complete step-by-step answer:
We have been given to prove \[1+{{i}^{10}}+{{i}^{20}}+{{i}^{30}}\] is a real number.
Now let us consider the case of \[{{i}^{10}}\] first. The power is 10, so it can be expressed in terms of multiple of 4 as 8 + 2. So, we can write it in the form of \[{{i}^{\left( 4\times 2+2 \right)}}\].
We know that,
\[\begin{align}
  & {{i}^{4n+2}}=-1 \\
 & {{i}^{10}}={{i}^{\left( 4\times 2+2 \right)}}=-1 \\
\end{align}\]
Again, we have the term \[{{i}^{20}}\]. We know that the power 20 is a multiple of 4. So, we can write it in the form of \[{{i}^{4\times 5}}\].
We know that,
\[\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{20}}={{i}^{4\times 5}}=1 \\
\end{align}\]
Now, we have the last term as, \[{{i}^{30}}\]. The power is 30, so it can be expressed in terms of multiple of 4 as 28 + 2. So, we can write it in the form of \[{{i}^{\left( 4\times 7+2 \right)}}\].
We know that,
\[\begin{align}
  & {{i}^{4n+2}}=-1 \\
 & {{i}^{30}}={{i}^{\left( 4\times 7+2 \right)}}=-1 \\
\end{align}\]
Hence we get the values of \[{{i}^{10}}=-1,{{i}^{20}}=1,{{i}^{30}}=-1\].
On substituting the values of \[{{i}^{10}},{{i}^{20}},{{i}^{30}}\] in the given expression, we get,
\[1+{{i}^{10}}+{{i}^{20}}+{{i}^{30}}=1+(-1)+(1)+(-1)=0\]
We know that 0 is a real number.
Therefore, it is proved that \[1+{{i}^{10}}+{{i}^{20}}+{{i}^{30}}\] is a real number.

Note: Be careful while calculating the value of \[{{i}^{10}},{{i}^{20}},{{i}^{30}}\]. Also remember that ‘i' is known as iota and it is equal to \[\sqrt{-1}\]. The most common mistake is thinking that the even power will be positive, but that is not the case here. So, we must be very careful in that regard.