Question

Find the modulus of $\dfrac{i+1}{1-i}$.

Answer 1

Hint: We must simplify and rationalize the given complex expression. And then we must use the formulae ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Then, we must use the formula for calculating the modulus, that is, $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.

Complete step-by-step solution:
We know that for any complex number $z=a+ib$, the modulus defines the distance of that point z from the origin. Since, modulus is a distance, it can never be a negative value.
We can define the modulus mathematically as,
$\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
Let us assume a complex variable $z=\dfrac{i+1}{1-i}$. We can also write this as
 $z=\dfrac{1+i}{1-i}$
Let us now multiply and divide the conjugate of the term present in the denominator, that is, 1 + i, on the right hand side of this equation.
Thus, we get
$z=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}$
We can write this as
\[z=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)}\]
We know that the expansion of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
So, we can write the equation as
\[z=\dfrac{1+{{i}^{2}}+2i}{\left( 1-i \right)\left( 1+i \right)}\]
We know that $i=\sqrt{-1}$ and thus, we can write that ${{i}^{2}}=-1$.
Thus, we get
\[z=\dfrac{1-1+2i}{\left( 1-i \right)\left( 1+i \right)}\]
Hence, we now have
\[z=\dfrac{2i}{\left( 1-i \right)\left( 1+i \right)}\]
We also know very well that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
With the help of this formula, we can write \[\left( 1-i \right)\left( 1+i \right)=1-{{i}^{2}}\]. But we know that the value of ${{i}^{2}}=-1$.
Thus, we can say that \[\left( 1-i \right)\left( 1+i \right)=1-\left( -1 \right)\]. Hence, we have \[\left( 1-i \right)\left( 1+i \right)=2\].
So, we can write the following equation comfortably,
\[z=\dfrac{2i}{2}\]
We can cancel the term 2 from both the numerator and denominator.
$z=i$
We can write the above equation once again in the following form
\[z=0+i1\]
Hence, the modulus of z can be written as
$\left| z \right|=\sqrt{{{0}^{2}}+{{1}^{2}}}$
Thus, we get $\left| z \right|=1$.
Hence, the modulus of $\dfrac{i+1}{1-i}$ is 1.

Note: We must remember that the modulus is always positive. So, we must not select the negative value that arises because of the square root, as the value of modulus. We must remember that during the rationalising process, we take the conjugate of the complex number in the denominator.