Question

Find the multiplicative inverse of \[\dfrac{\sqrt{3}}{2}-i\dfrac{1}{2}\] .

Answer 1

HINT: - Before solving this question, we must know about multiplicative inverse that is as follows:
In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by \[\dfrac{1}{x}\] or \[{{x}^{-1}}\] , is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth ( \[\dfrac{1}{5}\] or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4.

Complete step-by-step answer:
Now, in this question, we can find the multiplicative inverse by just letting the number as a+ib and then we will be able to get to the solution
As mentioned in the question, we have to find the multiplicative inverse of the given complex number.
Now, as mentioned in the hint, we can let the multiplicative inverse be a+ib as mentioned in the hint, so, now, we can write as follows to get to the answer
\[\begin{align}
  & \left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right)\cdot (a+ib)=1 \\
 & \dfrac{\sqrt{3}}{2}a-\dfrac{a}{2}i+\dfrac{\sqrt{3}}{2}bi-\dfrac{b}{2}{{i}^{2}}=1 \\
\end{align}\]
Now, we know that the value of \[{{i}^{2}}=-1\] , hence, we can write as follows
 \[\begin{align}
  & \dfrac{\sqrt{3}}{2}a-\dfrac{a}{2}i+\dfrac{\sqrt{3}}{2}bi-\dfrac{b}{2}(-1)=1 \\
 & \dfrac{\sqrt{3}}{2}a-\dfrac{a}{2}i+\dfrac{\sqrt{3}}{2}bi+\dfrac{b}{2}=1 \\
 & \left( \dfrac{\sqrt{3}}{2}a+\dfrac{b}{2} \right)+i\left( \dfrac{\sqrt{3}}{2}b-\dfrac{a}{2} \right)=1+0i \\
\end{align}\]
On comparing the real and the imaginary parts of the equation, we get the following result
\[\left( \dfrac{\sqrt{3}}{2}a+\dfrac{b}{2} \right)=1\ and\ \left( \dfrac{\sqrt{3}}{2}b-\dfrac{a}{2} \right)=0\ \ \ \ \ \ ...(a)\]
Now, using equation (a), we get
\[\begin{align}
  & \dfrac{\sqrt{3}}{2}b=\dfrac{a}{2} \\
 & a=\sqrt{3}b \\
\end{align}\]
Now, on using the above equation in (a), we get
\[\begin{align}
  & \left( \dfrac{\sqrt{3}}{2}a+\dfrac{b}{2} \right)=1 \\
 & \left( \dfrac{\left( \sqrt{3} \right)\left( b\sqrt{3} \right)}{2}+\dfrac{b}{2} \right)=1 \\
 & \left( \dfrac{\left( 3b \right)}{2}+\dfrac{b}{2} \right)=1 \\
 & \dfrac{4b}{2}=1 \\
 & b=\dfrac{1}{2} \\
\end{align}\]
Now, using the value of b to get the value of a as follows
\[\begin{align}
  & a=\sqrt{3}b \\
 & a=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Hence, the multiplicative inverse of the given complex number is as follows
\[a+ib=\dfrac{\sqrt{3}}{2}+i\dfrac{1}{2}\]

NOTE: - The students can make an error if they don’t know about the definition and the actual meaning of multiplicative inverse that is given in the hint as well because if one doesn’t know about all these things then he would not be able to get to the correct answer.
Also, in such questions, calculation mistakes are very common, so, to get to the right answer, one should try to be extra careful while solving these questions.
As calculated, the reciprocal of the given number \[\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right)\] is \[\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i \right)\] . So, if we will multiply the number and its reciprocal, then we should get 1.
\[\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right)\] \[\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i \right)\] = 1
\[\begin{align}
  & \left( \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} \right)i+\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}i-\dfrac{1}{2}\times \dfrac{1}{2}i_{{}}^{2}=1 \\
 & \dfrac{3}{4}-\dfrac{\sqrt{3}}{4}i+\dfrac{\sqrt{3}}{4}i-\dfrac{1}{4}i_{{}}^{2}=1 \\
 & 1=1 \\
\end{align}\]