Question

Find the multiplicative inverse of $\sqrt{5}+3i$

Answer 1

We solve this problem by using the definition of multiplicative inverse. A number $'x'$ is said to be the multiplicative inverse of $'n'$ if and only if
$n\times x=1$
Then we rationalise the fraction we get in the complex number by multiplying and dividing with the conjugate of the given complex number. The conjugate of a complex number $\left( x+iy \right)$ is given as $\left( x-iy \right)$ then, we use the standard value of imaginary number that is
$i=\sqrt{-1}$

Complete step by step answer:
We are given with the complex number that is $\sqrt{5}+3i$
Let us assume that the given number as
$\Rightarrow C=\sqrt{5}+3i$
Let us assume that the multiplicative inverse of given number as $Z$
We know that the number $'x'$ is said to be the multiplicative inverse of $'n'$ if and only if
$n\times x=1$
By using the above formula to given number we get
$\Rightarrow C\times Z=1$
Now, by substituting the required values in above equation we get

& \Rightarrow \left( \sqrt{5}+3i \right)\times Z=1 \\\ & \Rightarrow Z=\dfrac{1}{\sqrt{5}+3i}..........equation(i) \\\ \end{aligned}$$ Now, let us multiply and divide the RHS with the conjugate of $$\sqrt{5}+3i$$ We know that the conjugate of a complex number $$\left( x+iy \right)$$ is given as $$\left( x-iy \right)$$ By using the above formula we get the conjugate of $$\sqrt{5}+3i$$ as $$\sqrt{5}-3i$$ Now, by multiplying and dividing RHS of equation (i) with $$\sqrt{5}-3i$$ we get $$\begin{aligned} & \Rightarrow Z=\dfrac{1}{\sqrt{5}+3i}\left( \dfrac{\sqrt{5}-3i}{\sqrt{5}-3i} \right) \\\ & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{\left( \sqrt{5}+3i \right)\left( \sqrt{5}-3i \right)} \\\ \end{aligned}$$ We know that the standard formula of algebra that is $$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$$ By using this formula in above equation we get $$\Rightarrow Z=\dfrac{\sqrt{5}-3i}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( 3i \right)}^{2}}}$$ We know that the value of imaginary number that is $$\begin{aligned} & \Rightarrow i=\sqrt{-1} \\\ & \Rightarrow {{i}^{2}}=-1 \\\ \end{aligned}$$ By using this value of imaginary number in above equation we get $$\begin{aligned} & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{5-\left( -9 \right)} \\\ & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{14} \\\ \end{aligned}$$ **Therefore, the multiplicative inverse of $$\sqrt{5}+3i$$ is $$\dfrac{\sqrt{5}-3i}{14}$$.** **Note:** Students may leave the solution in the middle. Here, we have the multiplicative inverse of $$\sqrt{5}+3i$$ as $$\Rightarrow Z=\dfrac{1}{\sqrt{5}+3i}$$ Now, we need to rationalise the above number to get some other complex number. Therefore the answer will be $$\Rightarrow Z=\dfrac{\sqrt{5}-3i}{14}$$ Also, students may misunderstand the difference between the conjugate and multiplicative inverse of a complex number. The conjugate of a complex number $$\left( x+iy \right)$$ is given as $$\left( x-iy \right)$$ The number $$'x'$$ is said to be the multiplicative inverse of $$'n'$$ if and only if $$n\times x=1$$ These two parts need to be taken care of.