Question

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

Answer 1

Hint: 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 \[{{i}^{2}}=-1\].