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A complex number, (a + bi), multiplied by (2 + 3i) and added to -i gives the product of (-11 + 5i) and (1 – i).
a = and b =

Sagot :

lhannearaujo

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

Read more about complex numbers here:

https://brainly.com/question/5564133

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