Answer :
The value of a is equal to 3 and the value of b is equal to 4.
Complex Number
A Complex Number is represented by z=a+bi, where:
a= real part
bi= imaginary number
b= imaginary part
i= [tex]\sqrt{-1}[/tex], therefore i²= -1
For solving this question, you should solve the product (-11 + 5i) * (1 – i). After that, you should compare the previous results with the result for expression ( (a + bi)* (2 + 3i) )-i. From this comparison, using the linear system you will find the variables a and b.
- Step 1 - Solve the product (-11 + 5i) * (1 – i)
(-11 + 5i) * (1 – i)=
-11 +11i +5i -5i²
-11 +16i -5i², as i²= -1 you can rewrite
-11 +16i -5(-1)
-11 +16i +5
-6+16i
Then,
(-11 + 5i) * (1 – i)= -6+16i
- Step 2 - Simplify the expression ( (a + bi)* (2 + 3i) )-i
( (a + bi)* (2 + 3i) )-i
(2a+3ai +2bi +3bi²) -i, as i²= -1 you can rewrite
(2a+3ai +2bi +3b(-1)) -i
(2a+3ai +2bi -3b) -i
( (2a-3b) + i(3a +2b) ) -i ,
For adding complex numbers, you must add or subtract the corresponding real and imaginary parts. Thus, you will have:
(2a-3b) + i(3a +2b-1)
Then,
( (a + bi)* (2 + 3i) )-i= (2a-3b) + i(3a +2b-1)
- Step 3 - Identify real and imaginary parts for previous results
For (-11 + 5i) * (1 – i)= -6+16i
real part = -6
imaginary part = 16
For ( (a + bi)* (2 + 3i) )-i= (2a-3b) + i(3a +2b-1)
real part = 2a-3b
imaginary part = 3a +2b-1
- Step 4 - Construct a linear system from real and imaginary parts
2a-3b= -6
3a +2b-1=16, you can rewrite as 3a +2b=17.
Thus the system will be:
[tex]\begin{bmatrix}2a-3b=-6\\ 3a+2b=17\end{bmatrix}[/tex]
- Step 5 - Solving the linear system
2a-3b=-6 (1)
3a+2b=17 (2)
Multiplying equation 1 by 2 and equation 2 by 3, you can rewrite as:
4a-6b= -12
9a+6b=51
Sum the equations
4a-6b+9a+6b= -12 +51
13a= 39
a=39/13=3
Replacing the value of a=3 in equation 1, you can find b.
2a-3b=-6 (1)
2*3-3b= -6
6 -3b = -6
-3b = -6 -6
-3b= -12
b= -12/-3 = 4
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