Question

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

Answer 1

Hint: In complex number$z=x+iy$, the modulus of this complex number is$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Then simplify the expression given in the question and then the expression will be in the form of $z=x+iy$ then compare the values of x and y and put it in the modulus formula.

Complete step-by-step answer:
First of all we are going to simplify the expression given in the question.
$\begin{align}
  & \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i} \\
 & \Rightarrow \dfrac{\left( 1+i \right)\left( 1+i \right)-\left( 1-i \right)\left( 1-i \right)}{\left( 1-i \right)\left( 1+i \right)} \\
\end{align}$
In the above expression, ${(1 + i)}^{2}$ is in the form of ${(a + b)}^{2}$ and ${(1 – i)}^{2}$ is in the form of ${(a – b)}^{2}$ and $\left( 1+i \right)\left( 1-i \right)$is in the form of$\left( a+b \right)\left( a-b \right)$.
Applying these identities of $\left( a+b \right),\left( a-b \right),\left( a+b \right)\left( a-b \right)$in the above expression we get,
$\begin{align}
  & \Rightarrow \dfrac{1+2i+{{i}^{2}}-1+2i-{{i}^{2}}}{1-{{i}^{2}}} \\
 & \Rightarrow \dfrac{4i}{2} \\
 & \Rightarrow 2i \\
\end{align}$
In the above steps, we have used the value of ${i}^{2}$ = -1.
The simplification of the given expression is 2i.
Now, compare z = 2i with z = x +iy we get x = 0 and y = 2.
We know that modulus of z or$\left| z \right|$is$\sqrt{{{x}^{2}}+{{y}^{2}}}$.
So, $\left| 2i \right|=\sqrt{0+{{\left( 4 \right)}^{2}}}=4$
Hence, the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$is 4.

Note: Some interesting points about the complex number:
From basic algebra, modulus of any number is always positive. So, if you have given a multiple choice question then you can eliminate some options in which the answer is with a negative sign.
The modulus of a complex number is the length of a vector drawn in a complex plane. And we know that length cannot be negative.
The given expression simplifies to 2i which is a purely imaginary number. Purely imaginary means real part is 0 and in the complex plane real part lies on x axis and imaginary part lies on y axis so 2i lies on y axis.