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.