Question

How that $1+i{}^{10}+i^{20}+i^{30}$ is a real number.

Answer 1

Hint: The real numbers can be defined as the numbers which include natural numbers, whole numbers and integers. Here, we will use the concept of power to power rule and apply the value $i^2=-1$ for all the terms in the expression and then simplify for the required value.

Complete step-by-step answer:
Take the given expression: $1+i{}^{10}+i^{20}+i^{30}$
Now as per the imaginary value place $i^2=(-1)$
 $i^{10}=i^{2(5)}=(-1)$ …. (A)
And similarly, $i^{20}=i^{2(10)}=(-1)\times (-1)=1$ …. (B)
 [Product of two negative terms gives the resultant term in positive]
And similarly use the above value for the next term –
 $i^{30}=i^{3(10)}=(-1)\times (-1)\times (-1)=(-1)$ ….. (C)
Product of three negative terms gives the resultant term in negative by the fact that the product of the first two negative terms gives a positive term and its value multiplied with the third negative term gives a negative term.
Place the values of the equations (A), (B) and (C) in the given expression –
 $1+i{}^{10}+i^{20}+i^{30}$
 $=1+(-1)+1+(-1)$
Open the brackets, when there is a positive sign outside the bracket then the sign of the terms inside the bracket remains the same.
 $=1-1+1-1$
Terms with the same value and opposite sign cancel each other and therefore the term $+1$ and $(-1)$ cancel each other.
 $=0$
The resultant value is included in the set of real numbers and therefore the given expression $1+i{}^{10}+i^{20}+i^{30}$ is a real number.

Note: Always remember the concepts of the complex numbers which are the combinations of real numbers and the imaginary numbers and since imaginary numbers are very difficult to understand and therefore they are complex numbers. Always remember the value of imaginary “I” and accordingly find the ith power times value. Also, be good in multiples and simplifications of the equation. Remembering the square of the negative terms also gives the positive values.