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Question: Find the modulus of: \(\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\)...

Find the modulus of:
1+i1−i−1−i1+i\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}

Explanation

Solution

Hint: In complex numberz=x+iyz=x+iy, the modulus of this complex number is∣z∣=x2+y2\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+iyz=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.
1+i1−i−1−i1+i ⇒(1+i)(1+i)−(1−i)(1−i)(1−i)(1+i) \begin{aligned} & \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{aligned}
In the above expression, (1+i)2{(1 + i)}^{2} is in the form of (a+b)2{(a + b)}^{2} and (1–i)2{(1 – i)}^{2} is in the form of (a–b)2{(a – b)}^{2} and (1+i)(1−i)\left( 1+i \right)\left( 1-i \right)is in the form of(a+b)(a−b)\left( a+b \right)\left( a-b \right).
Applying these identities of (a+b),(a−b),(a+b)(a−b)\left( a+b \right),\left( a-b \right),\left( a+b \right)\left( a-b \right)in the above expression we get,
⇒1+2i+i2−1+2i−i21−i2 ⇒4i2 ⇒2i \begin{aligned} & \Rightarrow \dfrac{1+2i+{{i}^{2}}-1+2i-{{i}^{2}}}{1-{{i}^{2}}} \\\ & \Rightarrow \dfrac{4i}{2} \\\ & \Rightarrow 2i \\\ \end{aligned}
In the above steps, we have used the value of i2{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∣z∣\left| z \right|isx2+y2\sqrt{{{x}^{2}}+{{y}^{2}}}.
So, ∣2i∣=0+(4)2=4\left| 2i \right|=\sqrt{0+{{\left( 4 \right)}^{2}}}=4
Hence, the modulus of 1+i1−i−1−i1+i\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.