Solve Complex Numbers: Express (1/Z1)-(1/Z2) as a+bi

[aisha]

Let Z1 = 3-i
Z2=7+2i express (1/Z1)-(1/Z2) in form a+bi
SOMEONE pleasezzzzzz HELP ME! I don't have a clue as to how to do this
What do I do?
Where do I start?

 

[nolachrymose]

Have you ever rationalized the denominator of a fraction before? This is pretty much the same thing. For instance, when you have a fraction such as \frac{5}{\sqrt{3}+2}, you multiply the top and bottom by the conjugate, because you know a difference of squares will result in a rational number, because the square root of a rational squared is a rational. The same idea applies here: just keep in mind that i=\sqrt{-1}, and apply the same concept.

Hope that helps! :)

 

[Parth Dave]

Do you know what a "complex conjugate" is?

 

[aisha]

Do you know what a "complex conjugate" is?

Yes I do it is the opposite sign well when dividing you take the denominator and divide by the conjugate I know that but in this question I don't know what to do or what order to do it in please help me pleasezz

 

[BobG]

Have you ever rationalized the denominator of a fraction before? This is pretty much the same thing. For instance, when you have a fraction such as \frac{5}{\sqrt{3}+2}, you multiply the top and bottom by the conjugate, because you know a difference of squares will result in a rational number, because the square root of a rational squared is a rational. The same idea applies here: just keep in mind that i=\sqrt{-1}, and apply the same concept.

Hope that helps! :)

nolachrymose described it pretty well.

If you substituted i for the square root of 3,

\frac{5}{\sqrt{-1}+2},

and multiplied by the conjugate, you'd get:

\frac{5(\sqrt{-1}-2)}{-1 - 4} = 2 - \sqrt{-1}

The i is square root of negative one. If you have 3i and square it, you get -9. Other than keeping the negative signs straight, it's just like working with a square root.

 

[aisha]

Ive never rationalized the denominator of a fraction so I am a little lost I tried multiplying 1/3-i first by the conjugate and got the answer of 3+i/10 and then i did the same for z2 and got 7-2i/53 but if i subtract the two i don't get the right answer please show me how do i divide first subtract use conjugates or what?

 

[Parth Dave]

As far as I can tell you rationalized the two properly. Unless you made a subtraction error, your answer should be right. What did you get as the final answer?

 

[aisha]

There are 4 possible answers
a)89/24-73/24i

b)89/24+73/24i

c)-89/24+73/24i

d)89/24i-73/24

I don't get anything close to these when i take those two answers and subtract them. Please help me I've been doing this one forever

 

[BobG]

All of their choices are wrong.

So what did you get? Something close to

\frac{89+73i}{530}

 

[BobG]

I think I know where they made their mistake. You can do this two ways.

\frac{1}{3-i}-\frac{1}{7+2i}

\frac{3+i}{9-(-1)}-\frac{7-2i}{49-(-4)}

\frac{3+i}{10}-\frac{7-2i}{53}

\frac{(159+53i)-(70-20i)}{530}

\frac{89+73i}{530}

Or:

\frac{1}{3-i}-\frac{1}{7+2i}

\frac{7+2i}{(3-i)(7+2i)}-\frac{3-i}{(3-i)(7+2i)}

\frac{7+2i-3+i}{(21-(-2))+(6i-7i)}

\frac{4+3i}{23-i}

\frac{(4+3i)(23+i)}{(23-i)(23+i)}

\frac{(92+(-3))+(69i+4i)}{529-(-1)} Here's where they made their mistake

\frac{89+73i}{530}

You can break this up into two separate fractions, if you want:

\frac{89}{530}+\frac{73}{530}i

Instead of multiplying (23-i)(23+i) and getting 529+1, they got 23+1. So, the answer they most likely picked is (b)

 

[arildno]

A great example of why multiple choice exams are (IMO) inferior to some other exam types which are less sensitive to designer flaws.

 

[BobG]

A great example of why multiple choice exams are (IMO) inferior to some other exam types which are less sensitive to designer flaws.

A flaw?! Or is it really a hidden 'feature'? (how come we don't have a 'shifty eyed' smilie?)

"B's for everyone who answers all the questions right. A's for everyone who catches my error and figures out exactly where I made my mistake!"

 

[aisha]

All of their choices are wrong.

So what did you get? Something close to

\frac{89+73i}{530}

Well my answer was (89-33i)/530
u are right that the question that was correct was b with the error in it. I think our answers are different because when u expanded -10(7-2i)/530 u wrote -70-20i shouldn't this be -70+2i because of the two negative signs? I will for sure point this question out to my teacher maybe she will give me an "A" lol for cathching the errors. Thanks so much I thought I was doing this totally wrong, but I guess not thanks again ur my HERO