Parviz F. answered • 09/26/13

Mathematics professor at Community Colleges

^{2 }* ( 3 - 2i ) /( 5 - i ) =

Parviz F.

09/26/13

Addie K.

asked • 09/25/13Write 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)

(3-2i)^2 (times) (3+2i) all over (5-i)

So the question is:

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

------------------

(5-i)

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

------------------

(5-i)

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Parviz F. answered • 09/26/13

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Mathematics professor at Community Colleges

( 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

Parviz F.

Not really. It just happened that I copied it wrong in scrolling the screen.

Giving the exact answer is not my problem. I always emphasize the fact that journey is more important than the destination. Student's problem is mostly with journey, because they don't follow steps correctly, and most of the time they forged the answer.

So, it is up to them to follow the steps with understanding each step. The way I write, shows them that how do I get from each step to next. Sometimes I do things in my head, and let them follow to find how did I get something.

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09/26/13

Susan L. answered • 09/25/13

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Tutor in Math and Science--Patient, Proficient, Organized

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.

Susan L.

Final should be 17/**2 - **(7/2)i

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09/25/13

Adrienne J. answered • 09/25/13

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Effective Math Tutor Specializing in Algebra

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.

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

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

Distribute the 13

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

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

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) **

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!

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Susan L.

09/26/13