Question

If we have a complex number as \[z=1+i\], then the multiplicative inverse of \[{{z}^{2}}\] is (where, \[i=\sqrt{-1}\])
a). \[2i\]
b). \[1-i\]
c). \[\dfrac{-i}{2}\]
d). \[\dfrac{i}{2}\]

Answer 1

Hint: First we will find \[{{z}^{2}}\], by finding the square of the given complex number, \[z=1+i\]. Then, we will find the multiplicative inverse. Multiplicative inverse is the number which, when multiplied to the original number gives the result as 1 which is the multiplicative identity. So we will assume the multiplicative inverse to be \[x+iy\] and multiply it with the original number. The result is equated to 1 and there we will get 2 conditions in x and y. Solving them would give us the final answer.

Complete step-by-step solution -
Let the multiplicative inverse of the complex number, \[{{z}^{2}}={{\left( 1+i \right)}^{2}}\] be \[x+iy\].
Now, let us multiply these two numbers.
Doing so we get, \[{{\left( 1+i \right)}^{2}}\left( x+iy \right)\]. This product is 1.
Thus, \[{{\left( 1+i \right)}^{2}}\left( x+iy \right)=1\]
Now, expanding the first term using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] formula, we have,
\[\begin{align}
  & \left( {{1}^{2}}+2\times 1\times i+{{i}^{2}} \right)\left( x+iy \right)=1 \\
 & \Rightarrow \left( 1+2i+{{i}^{2}} \right)\left( x+iy \right)=1 \\
\end{align}\]
Now, since \[i=\sqrt{-1}\], \[{{i}^{2}}=-1\].
Thus, \[\left( 1+2i-1 \right)\left( x+iy \right)=1\]
\[\Rightarrow \left( 2i \right)\left( x+iy \right)=1\]
Multiplying the two numbers we get,
\[\begin{align}
  & 2xi+\left( 2i\times yi \right)=1 \\
 & \Rightarrow \left( 2x \right)i+2{{i}^{2}}y=1 \\
\end{align}\]
Since, \[{{i}^{2}}=-1\] we get,
\[\begin{align}
  & \Rightarrow \left( 2x \right)i+2\left( -1 \right)y=1 \\
 & \Rightarrow \left( 2x \right)i-2y=1 \\
\end{align}\]
Now, 1 can be written as \[1+0i\].
Thus, \[-2y+\left( 2x \right)i=1+0i\]
Comparing the LHS and RHS, we have,
\[-2y=1\] and \[2x=0\]
Thus, \[y=\dfrac{-1}{2}\] and \[x=0\]
Thus, the number \[x+iy\] is \[0+i\left( \dfrac{-1}{2} \right)\], which is \[\left( \dfrac{-i}{2} \right)\].
Therefore, option (c) is the correct answer.

Note: This problem can be solved in a different way also. Let the multiplicative inverse of \[{{\left( 1+i \right)}^{2}}\] be \[\left( x+iy \right)\]. Thus multiplying the two numbers we get 1.
\[\Rightarrow {{\left( 1+i \right)}^{2}}\left( x+iy \right)=1\]
Expanding the first term using, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get,
\[\Rightarrow \left( {{1}^{2}}+2\times 1\times i+{{i}^{2}} \right)\left( x+iy \right)=1\]
Since, \[{{i}^{2}}=-1\], we get,
\[\begin{align}
  & \Rightarrow \left( 1+2i-1 \right)\left( x+iy \right)=1 \\
 & \Rightarrow \left( 2i \right)\left( x+iy \right)=1 \\
\end{align}\]
Cross multiplying \[2i\], we get,
\[\Rightarrow x+iy=\dfrac{1}{2i}\]
Now multiplying the RHS, by the conjugate of \[2i\], which is \[-2i\], we get,
\[\begin{align}
  & \Rightarrow x+iy=\dfrac{1}{2i}\times \dfrac{-2i}{-2i} \\
 & \Rightarrow x+iy=\dfrac{-2i}{-4{{i}^{2}}} \\
\end{align}\]
Since, \[{{i}^{2}}=-1\], we get,
\[\begin{align}
  & \Rightarrow x+iy=\dfrac{-2i}{-4\left( -1 \right)} \\
 & \Rightarrow x+iy=\dfrac{-2i}{4} \\
\end{align}\]
Simplifying, we get,
\[\Rightarrow x+iy=\left( \dfrac{-i}{2} \right)\]
Thus the multiplicative inverse is \[\left( \dfrac{-i}{2} \right)\].