Question

If ${z_1} = 2 - i$ any ${z_2} = 1 + i$ , then what is the value of $\left| {\dfrac{{\left( {{z_1} - {z_2} + 1} \right)}}{{\left( {{z_1} + {z_2} - i} \right)}}} \right|$ ?
A) $\sqrt {\dfrac{5}{3}}$
B) $\sqrt {\dfrac{3}{5}}$
C) $\sqrt {\dfrac{4}{5}}$
D) $\sqrt {\dfrac{5}{4}}$

Answer 1

Hint : In this question we have given two complex numbers ${z_1} = 2 - i$ any ${z_2} = 1 + i$ and we have to find the value of $\left| {\dfrac{{\left( {{z_1} - {z_2} + 1} \right)}}{{\left( {{z_1} + {z_2} - i} \right)}}} \right|$
The complex number is written in the form of $a + ib$ , where $a$ and $b$ are real numbers and $i$ is iota. In $z = a + ib$ , $a$ is called the real part and $b$ is called the imaginary part. To solve this question will put the value of ${z_1}$ and ${z_2}$ in $\left| {\dfrac{{\left( {{z_1} - {z_2} + 1} \right)}}{{\left( {{z_1} + {z_2} - i} \right)}}} \right|$ and after that we will find the value of modulus to get the final answer.
Let $z = a + ib$ be a complex number. Then, the modulus of $z$ , denoted by $\left| z \right|$ , is defined to be the non-negative real number $\sqrt {{a^2} + {b^2}}$ , i.e., $\left| z \right| = \sqrt {{a^2} + {b^2}}$ .

Complete step-by-step answer :
We have,
$\left| {\dfrac{{\left( {{z_1} - {z_2} + 1} \right)}}{{\left( {{z_1} + {z_2} - i} \right)}}} \right|$
Put the value of ${z_1} = 2 - i$ and ${z_2} = 1 + i$ .
$\Rightarrow \left| {\dfrac{{2 - i - \left( {1 + i} \right) + 1}}{{2 - i + 1 + i - i}}} \right|$
Open the bracket and change the sign.
$\Rightarrow \left| {\dfrac{{2 - i - 1 - i + 1}}{{2 - i + 1 + i - i}}} \right|$
$\Rightarrow \left| {\dfrac{{2 - 2i}}{{3 - i}}} \right|$
We can also write it as:
$\Rightarrow \dfrac{{\left| {2 - 2i} \right|}}{{\left| {3 - i} \right|}}$ …(1)
Now, we will solve $\left| {2 - 2i} \right|$ and $\left| {3 - i} \right|$ separately
$\left| {2 - 2i} \right| = \sqrt {{2^2} + {{\left( { - 2} \right)}^2}}$
$\left| {2 - 2i} \right| = \sqrt 8$
And,
$\left| {3 - i} \right| = \sqrt {{3^2} + {{\left( { - 1} \right)}^2}}$
$\left| {3 - i} \right| = \sqrt {9 + 1}$
Substitute these values in equation (1)
$\dfrac{{\left| {2 - 2i} \right|}}{{\left| {3 - i} \right|}} = \dfrac{{\sqrt 8 }}{{\sqrt {10} }}$
We can write it as:
$\dfrac{{\left| {2 - 2i} \right|}}{{\left| {3 - i} \right|}} = \dfrac{{\sqrt {4 \times 2} }}{{\sqrt {5 \times 2} }}$
On cancelling $\sqrt 2$ , we get
$\therefore \dfrac{{\left| {2 - 2i} \right|}}{{\left| {3 - i} \right|}} = \dfrac{{\sqrt 4 }}{{\sqrt 5 }}$
Hence, the correct option is 4.
So, the correct answer is “Option 4”.

Note : As in the above question $i$ i.e., iota is given, we can’t put the value of $i$ here because the value of $i$ is $\sqrt { - 1}$ . If we substitute the value of $i$ here, it will make the question more complex. The distance of the complex number represented as a point in the argand plane $\left( {a,ib} \right)$ is called the modulus of the complex number, so it can’t be negative. Here, modulus plays the most important role. Don’t just remove the modulus, solve the modulus by $\left| z \right| = \sqrt {{a^2} + {b^2}}$ formula.