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,

& \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{aligned}$$ Now, since $$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{aligned} & 2xi+\left( 2i\times yi \right)=1 \\\ & \Rightarrow \left( 2x \right)i+2{{i}^{2}}y=1 \\\ \end{aligned}$$ Since, $${{i}^{2}}=-1$$ we get, $$\begin{aligned} & \Rightarrow \left( 2x \right)i+2\left( -1 \right)y=1 \\\ & \Rightarrow \left( 2x \right)i-2y=1 \\\ \end{aligned}$$ Now, 1 can be written as $$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{aligned} & \Rightarrow \left( 1+2i-1 \right)\left( x+iy \right)=1 \\\ & \Rightarrow \left( 2i \right)\left( x+iy \right)=1 \\\ \end{aligned}$$ Cross multiplying $$2i$$, we get, $$\Rightarrow x+iy=\dfrac{1}{2i}$$ Now multiplying the RHS, by the conjugate of $$2i$$, which is $$-2i$$, we get, $$\begin{aligned} & \Rightarrow x+iy=\dfrac{1}{2i}\times \dfrac{-2i}{-2i} \\\ & \Rightarrow x+iy=\dfrac{-2i}{-4{{i}^{2}}} \\\ \end{aligned}$$ Since, $${{i}^{2}}=-1$$, we get, $$\begin{aligned} & \Rightarrow x+iy=\dfrac{-2i}{-4\left( -1 \right)} \\\ & \Rightarrow x+iy=\dfrac{-2i}{4} \\\ \end{aligned}$$ Simplifying, we get, $$\Rightarrow x+iy=\left( \dfrac{-i}{2} \right)$$ Thus the multiplicative inverse is $$\left( \dfrac{-i}{2} \right)$$.