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

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Homework Help Overview

The discussion revolves around expressing the difference of the reciprocals of two complex numbers, Z1 and Z2, in the form a + bi. The original poster expresses confusion about how to approach the problem involving complex numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rationalizing denominators and the use of complex conjugates. Some express uncertainty about the order of operations and the steps needed to simplify the expression. Others share their attempts at rationalization and subtraction of the resulting fractions.

Discussion Status

There is ongoing exploration of different methods to solve the problem, with participants sharing their calculations and questioning the correctness of their results. Some guidance has been provided regarding the use of conjugates, but no consensus has been reached on the final answer.

Contextual Notes

Participants mention potential errors in calculations and the challenges posed by multiple-choice options that may not align with their results. There is also a reference to the complexity of the problem due to the nature of complex numbers.

aisha
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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 :cry:
What do I do?
Where do I start? :cry:
 
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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 [tex]\frac{5}{\sqrt{3}+2}[/tex], 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 [tex]i=\sqrt{-1}[/tex], and apply the same concept.

Hope that helps! :)
 
Do you know what a "complex conjugate" is?
 
Parth Dave said:
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
 
nolachrymose said:
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 [tex]\frac{5}{\sqrt{3}+2}[/tex], 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 [tex]i=\sqrt{-1}[/tex], and apply the same concept.

Hope that helps! :)
nolachrymose described it pretty well.

If you substituted i for the square root of 3,

[tex]\frac{5}{\sqrt{-1}+2}[/tex],

and multiplied by the conjugate, you'd get:

[tex]\frac{5(\sqrt{-1}-2)}{-1 - 4} = 2 - \sqrt{-1}[/tex]

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.
 
Ive never rationalized the denominator of a fraction :frown: 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?
 
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?
 
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 :mad:
 
All of their choices are wrong.

So what did you get? Something close to

[tex]\frac{89+73i}{530}[/tex]
 
  • #10
I think I know where they made their mistake. You can do this two ways.

[tex]\frac{1}{3-i}-\frac{1}{7+2i}[/tex]

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

[tex]\frac{3+i}{10}-\frac{7-2i}{53}[/tex]

[tex]\frac{(159+53i)-(70-20i)}{530}[/tex]

[tex]\frac{89+73i}{530}[/tex]

Or:

[tex]\frac{1}{3-i}-\frac{1}{7+2i}[/tex]

[tex]\frac{7+2i}{(3-i)(7+2i)}-\frac{3-i}{(3-i)(7+2i)}[/tex]

[tex]\frac{7+2i-3+i}{(21-(-2))+(6i-7i)}[/tex]

[tex]\frac{4+3i}{23-i}[/tex]

[tex]\frac{(4+3i)(23+i)}{(23-i)(23+i)}[/tex]

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

[tex]\frac{89+73i}{530}[/tex]

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

[tex]\frac{89}{530}+\frac{73}{530}i[/tex]

Instead of multiplying (23-i)(23+i) and getting 529+1, they got 23+1. So, the answer they most likely picked is (b)
 
  • #11
A great example of why multiple choice exams are (IMO) inferior to some other exam types which are less sensitive to designer flaws.
 
  • #12
arildno said:
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'? :devil: (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!"
 
  • #13
BobG said:
All of their choices are wrong.

So what did you get? Something close to

[tex]\frac{89+73i}{530}[/tex]

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 :smile:
 

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