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Question: Let \[z=\dfrac{11-3i}{1+i}\]. If a is the real number such that \[z-ia\] is real then the value of a...

Let z=113i1+iz=\dfrac{11-3i}{1+i}. If a is the real number such that ziaz-ia is real then the value of a is
A) 11
B) 4-4
C) 77
D) 7-7

Explanation

Solution

In this type of problem, it has been mentioned the value of z=113i1+iz=\dfrac{11-3i}{1+i} from this value we have to rationalize the denominator so that we get the value and in question it is mentioned that ziaz-ia is real that means the imaginary part is zero. The value of the imaginary part which we get by rationalizing that we have to compare and equate it to zero (imaginary is zero) to get the value of a.

Complete step by step solution:
Here, you can notice that expression is given z=113i1+iz=\dfrac{11-3i}{1+i}
And we have to simplify the expression by rationalizing the denominator.
z=(113i)(1+i)×(1i)(1i)z=\dfrac{\left( 11-3i \right)}{\left( 1+i \right)}\times \dfrac{\left( 1-i \right)}{\left( 1-i \right)}
By simplifying the above expression we get:
z=(113i)(1i)(1+i)(1i)z=\dfrac{\left( 11-3i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)}
Now, if you observe in the denominator then it is in the form of (a+b)(ab)\left( a+b \right)\left( a-b \right)
So, we have to apply the property of above form that is (a+b)(ab)=a2b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} substitute this property in above expression we get
z=1111i3i+3i21i2z=\dfrac{11-11i-3i+3{{i}^{2}}}{1-{{i}^{2}}}
As we know that i2=1{{i}^{2}}=-1 substitute in this above expression we get:
z=1111i3i+3(1)1(1)z=\dfrac{11-11i-3i+3(-1)}{1-(-1)}
Further simplifying the above expression we get:
z=814i2z=\dfrac{8-14i}{2}
Split this above expression into two part that is real part as well as imaginary part us get:
z=47i(1)z=4-7i---(1)
Now, another equation which is given that is ziaz-ia if you observe then we have to find the value of a
For that we have to add ia-ia on both sides on equation (1) then, we get:
zia=47iiaz-ia=4-7i-ia
On RHS separate the term as real and imaginary part we get:
zia=4i(7+a)z-ia=4-i(7+a)
If you observe ziaz-ia in the question then it has been mentioned clearly that ziaz-ia is real, which means the imaginary part must be zero.
zia=4i(7+a)z-ia=4-i(7+a)
7+a=07+a=0
By simplifying further we get:
a=7a=-7
Therefore, the correct option is option (D).

Note: In this type of problem always remember that first of all we have to rationalize the given expression after that, separate the term in real and imaginary, modify the expression according to condition and by comparing we get the value of a. Most of the students make mistakes while comparing with ziaz-ia and write the value of aa is 7 but that is wrong because if you carefully see in this question it has written that ziaz-ia is real that means the imaginary part is zero. So we have to equate the expression of the imaginary part with zero then we get the value of a.