Complex number in standard form?

Question

Write the complex number in standard form:

 

the problem is actually (3-2i) to the second power:
(3-2i)^2  (times) (3+2i) all over (5-i)

So the question is:

(3-2i)^2 (3+2i)
------------------
          (5-i)

Answer 1

  

 

( 3-2i )² * ( 3 - 2i ) /( 5 - i ) = 

 

( 9 + 4 - 6i ) ( 3 - 2i ) / ( 5 - i) =

 

 ( 39 -  18i -26i-12) /(5 -i) =

 

( 27 - 44i ) /(5 - i ) =

 

(27 -44i ) ( 5 + i ) / ( 5 -i) ( 5 + i) =

 

  ( 135 - 220i +27i+44) / 26

 

  179/26 +7/26 i  

Answer 2

A complex number in standard form is a + bi

 

For multiplying and dividing complex numbers remember that i^2 = -1 and when you multiply a complex number (a +bi) by its conjugate (a - bi) the resulting product is always a real number and positive.

 

For this problem we can first factor out a conjugate and its pair and multiply.

 

(3-2i)^2 (3+2i)/(5-i)

 

= (3-2i)(3-2i)(3+2i)/(5-i)

 

=((3-2i) ((3(3+2i) - 2i(3+2i))/(5-i)

 

=(3-2i)(9+6i-6i-4i^2)      (3-2i)13 

   ----------------------  = ----------

           5-i                         5-i             (remember i^2 = -1)

 

To divide complex numbers we will rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator.

 

13 (3-2i)    5+i

---------- x -----

  5-i            5+i

 

= 13((3(5+i) -2i(5+i)) 

   -----------------------

        5(5+i) - i(5+i) 

 

= 13(15+3i-10i-2i^2)

   ----------------------- 

            25+5i-5i-i^2

 

= 13(15-7i+2) 

   --------------

       25+1

 

= 13(17-7i)

   -----------

        26

 

= (17-7i)/2 = 17/7 - (7/2)i

 

 Sorry it looks like my margins are different when I submit answer so it is not lining up and I don't have same screen to fix it. I will try to break up on different lines.

Answer 3

In order to get a complex number into standard form, you must not have any complex numbers in the denominator.

 

So first we must simplify the problem. For clarity, I am going to write out the problem.

 

(3-2i)(3-2i)(3+2i)

_______________     

          (5-i)

 

Whenever you are multiplying problems that have (a-bi)*(a+bi), you can use the shortcut (a^2-((b^2)(i^2))).

 

This is called a difference of squares.

 

Therefore we now have

 

(3-2i)(3-2i)(3+2i)         (3-2i)(3+2i)(3-2i)

_______________   =  _______________

      (5-i)                                (5-i)  

 

 

 

      (9-4i^2)(3-2i)

  = ____________

           (5-i)

 

 

Since i^2=-1, we can simplify this to be

 

   (9-(-4))(3-2i)

=___________

         (5-i)

 

 

    (9+4)(3-2i)

= ___________

         (5-i)

 

= (13)(3-2i)

____________

         (5-i)

 

Distribute the 13

 

    (39-26i)

= ________

       (5-i)

 

Now we need to get this answer into standard form, so we are going to use the difference of squares method and multiply by the fraction

 

(5+i)

_____

(5+i)         

 

We are able to do this because the fraction above is equal to one, and multiplying any number by one gives us the same number.

 

So now we have

 

      (39-26i)      (5+i)

= ________ *  ______

        (5-i)          (5+i)

 

Using the FOIL method you can simplify the numerator and using the difference of squares method for the denominator you should have

 

DENOMINATOR: (5-i)(5+i) = (25-i^2)

                                       = (25-(-1))

                                       = (25+1)

                                       = 26

 

(39-26i)(5+i)

___________   

        (26)

 

So, once you FOIL the top, what are you left with?

 

Try working this problem with the suggestions I gave you and see if you can simplify it to Standard form.

 

Let me know if you have any further questions!