Question

How do you find the multiplicative inverse of 5+2i in standard form?

Answer 1

Hint: In this question, we are given a complex number and we need to find its multiplicative inverse in the standard form. The standard form of complex numbers is a+ib. We will first find multiplicative inverse as $\dfrac{1}{5+2i}$ and then convert it into standard form by using multiplication and division by a certain number a-ib. After simplifying it, we will get the required inverse in standard form. We will use property that ${{i}^{2}}=-1\text{ and }\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.

Complete step by step answer:
Here we are given the complex number as 5+2i. We need to find the multiplicative inverse of 5+2i in standard form. For this let us first understand the meaning of multiplicative inverse.
A number u is said to be a multiplicative inverse of v if the product of u with v gives the answer 1. Here we are given 5+2i and we need to find a number such that (5+2i)v = 1. As we can see, if we take v as $\dfrac{1}{5+2i}$ then equality holds and thus we have the multiplicative inverse as $\dfrac{1}{5+2i}$. But we need it in the standard form i.e. in the form a+ib. For this let us multiply and divide the number by the conjugate of the denominator.
Conjugate of a+ib is given as a-ib.
Here the denominator is 5+2i. So its conjugate will be 5-2i. Therefore, we will multiply and divide $\dfrac{1}{5+2i}$ by 5-2i. We get \[\dfrac{1}{5+2i}\times \dfrac{5-2i}{5-2i}\].
It can be written as \[\dfrac{5-2i}{\left( 5+2i \right)\left( 5-2i \right)}\].
We can see that the denominator of the form (a+b)(a-b). So we can use the arithmetic property that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Hence taking 5 and 2i and applying it on denominator we get \[\dfrac{5-2i}{{{\left( 5 \right)}^{2}}-{{\left( 2i \right)}^{2}}}\].
We know that \[{{\left( 5 \right)}^{2}}=5\times 5=25\text{ and }{{\left( 2i \right)}^{2}}\] can be written as \[{{\left( 2i \right)}^{2}}{{i}^{2}}\] where \[{{\left( 2i \right)}^{2}}=2\times 2=4\] so we get \[\dfrac{5-2i}{25-4{{i}^{2}}}\].
Now we know that, value of i is equal to $\sqrt{-1}$ i.e. $i=\sqrt{-1}$. Squaring both sides we get ${{i}^{2}}=-1$. Therefore, the expression becomes \[\dfrac{5-2i}{25-\left( -4 \right)}\].
Simplifying the denominator we have \[\dfrac{5-2i}{25+4}\Rightarrow \dfrac{5-2i}{29}\].
Separating the terms from the numerator we get \[\dfrac{5}{29}-\dfrac{2}{29}i\].
As we can see the number is of the form a+ib where \[a=\dfrac{5}{29}\text{ and }b=\dfrac{2}{29}\]. Therefore, it is in standard form.
Hence the multiplicative inverse of 5+2i is \[\dfrac{5}{29}-\dfrac{2}{29}i\].

Note:
Students should note that, for any complex number z its multiplicative inverse is $\dfrac{1}{z}$. While using (a+b)(a-b) take care of the signs and do not forget to take the square of i as well. Lastly, always separate the terms using denominator separately with terms of numerator.