Question

Let \[z = \dfrac{{\left( {11 - 3i} \right)}}{{\left( {1 + i} \right)}}\] . If \[a\] is a real number such that \[z - ia\] is real, then what is the value of \[a\] is?
A. 4
B. -4
C. 7
D. -7

Answer 1

Hint In the question, the given equation is a complex equation. We will simplify the equation by multiplying the numerator and denominator by the conjugate of the denominator. By substituting the value of \[z\] in the equation \[z - ia\] and comparing the imaginary part with zero, we will find the value of \[a\].

Formula used:
The conjugate of a complex number \[z = a + ib\] is \[\overline z = a - ib\].
The product of a complex number and its conjugate is:
\[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\]

Complete step by step solution:
The given equation is \[z = \dfrac{{\left( {11 - 3i} \right)}}{{\left( {1 + i} \right)}}\].
Let’s multiply the numerator and denominator of the equation by the conjugate of the denominator.
The conjugate of the denominator \[\left( {1 + i} \right)\] is \[\left( {1 - i} \right)\].
Multiply the numerator and denominator by \[\left( {1 - i} \right)\].
\[z = \dfrac{{\left( {11 - 3i} \right)\left( {1 - i} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}}\]
Simplify the above equation.
Apply the formula \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\] for the denominator.
\[z = \dfrac{{11 - 11i - 3i + 3{i^2}}}{{\left( {{1^2} + {1^2}} \right)}}\]
\[ \Rightarrow \]\[z = \dfrac{{11 - 14i + 3\left( { - 1} \right)}}{{\left( {1 + 1} \right)}}\] [since \[{i^2} = - 1\]]
\[ \Rightarrow \]\[z = \dfrac{{8 - 14i}}{2}\]
\[ \Rightarrow \]\[z = 4 - 7i\]
Now substitute \[z = 4 - 7i\] in the expression\[z - ia\].
\[z - ia = 4 - 7i - ia\]
\[ \Rightarrow \]\[z - ia = 4 - \left( {7 + a} \right)i\]
It is given that \[z - ia\] is a real number.
This means the imaginary part is equal to zero.
The imaginary part of \[z - ia\] is \[- 7 - a\].
So,
\[- 7 - a = 0\]
\[ \Rightarrow \] \[a = - 7\]
Hence the correct option is option D.

Note: Students are often confused with the formulas \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\] and \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} - {b^2}} \right)\] .
 But, the correct formula of the product of a complex number and its conjugate is \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\]. Because the square of the imaginary number \[i\] is - 1, which is multiplied by \[ - {b^2}\].