Question

How do you simplify the expression ${{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}$?

Answer 1

In order to find the solution of the given question that is to simplify ${{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}$ use the one of the well-known properties of iota in complex numbers that is ${{i}^{2}}=-1$ to simplify the given expression and also use the result that ${{i}^{4n}}=1$ where n is any natural number.

Complete step-by-step solution:
According to the question, given expression in the question is as follows:
${{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}$
We will solve each five term individually then substitute its value in the given expression, so first we will solve ${{i}^{44}}$, here we will have:
$\Rightarrow {{i}^{44}}={{\left( {{i}^{4}} \right)}^{11}}$
We will use the following the property of iota in complex number that is ${{i}^{4n}}=1$ where n is any natural number in the above equation, we will get:
$\Rightarrow {{i}^{44}}={{\left( 1 \right)}^{11}}$
$\Rightarrow {{i}^{44}}=1$
Now we will solve for ${{i}^{150}}$, we will have:
$\Rightarrow {{i}^{150}}={{\left( {{i}^{4}} \right)}^{37}}\cdot {{i}^{2}}$
To simplify the above equation, we will use the following two properties of iota in complex number that is ${{i}^{2}}=-1$ and ${{i}^{4n}}=1$ where n is any natural number, we get:
$\Rightarrow {{i}^{150}}={{\left( 1 \right)}^{37}}\cdot {{i}^{2}}$
$\Rightarrow {{i}^{150}}=1\cdot {{i}^{2}}$
$\Rightarrow {{i}^{150}}=-1$
After this we will solve for ${{i}^{74}}$, we will have:
$\Rightarrow {{i}^{74}}={{\left( {{i}^{4}} \right)}^{18}}\cdot {{i}^{2}}$
To simplify the above equation, we will use the following two properties of iota in complex number that is ${{i}^{2}}=-1$ and ${{i}^{4n}}=1$ where n is any natural number, we get:
$\Rightarrow {{i}^{74}}={{\left( 1 \right)}^{18}}\cdot {{i}^{2}}$
$\Rightarrow {{i}^{74}}=1\cdot {{i}^{2}}$
$\Rightarrow {{i}^{74}}=-1$
Now solve for the term ${{i}^{109}}$, we will have:
$\Rightarrow {{i}^{109}}={{\left( {{i}^{4}} \right)}^{27}}\cdot i$
We will use the following the property of iota in complex number that is ${{i}^{4n}}=1$ where n is any natural number in the above equation, we will get:
$\Rightarrow {{i}^{109}}={{\left( 1 \right)}^{27}}\cdot i$
$\Rightarrow {{i}^{109}}=1\cdot i$
$\Rightarrow {{i}^{109}}=i$
At last, we will solve for ${{i}^{61}}$, we get:
$\Rightarrow {{i}^{61}}={{\left( {{i}^{4}} \right)}^{15}}\cdot i$
We will use the following the property of iota in complex number that is ${{i}^{4n}}=1$ where n is any natural number in the above equation, we will get:
$\Rightarrow {{i}^{61}}={{\left( 1 \right)}^{15}}\cdot i$
$\Rightarrow {{i}^{61}}=1\cdot i$
$\Rightarrow {{i}^{61}}=i$
Now substitute the values of all the five terms in the given expression, we will get:
$\Rightarrow {{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}=1+\left( -1 \right)-\left( -1 \right)-i+i$
$\Rightarrow {{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}=1-1+1-i+i$
$\Rightarrow {{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}=1$
Therefore, the value of the expression ${{i}^{44}}+{{i}^{150}}-{{i}^{74}}-{{i}^{109}}+{{i}^{61}}$ is $1$.

Note: Students make mistakes because using the wrong property of iota in complex numbers, that is, they use $i=-1$ which is completely incorrect and leads to the wrong answer. It’s important to remember that ${{i}^{2}}=-1$.