Question

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

Answer 1

Hint: 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
\[\begin{align}
  & \Rightarrow \left( \sqrt{5}+3i \right)\times Z=1 \\
 & \Rightarrow Z=\dfrac{1}{\sqrt{5}+3i}..........equation(i) \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow Z=\dfrac{1}{\sqrt{5}+3i}\left( \dfrac{\sqrt{5}-3i}{\sqrt{5}-3i} \right) \\
 & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{\left( \sqrt{5}+3i \right)\left( \sqrt{5}-3i \right)} \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
\end{align}\]
By using this value of imaginary number in above equation we get
\[\begin{align}
  & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{5-\left( -9 \right)} \\
 & \Rightarrow Z=\dfrac{\sqrt{5}-3i}{14} \\
\end{align}\]

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.