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

In summary, someone is trying to solve a fraction problem, but is lost and cannot seem to get the answer correct. They try two different methods and both of them end in error.
  • #1
aisha
584
0
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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  • #2
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! :)
 
  • #3
Do you know what a "complex conjugate" is?
 
  • #4
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
 
  • #5
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.
 
  • #6
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?
 
  • #7
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?
 
  • #8
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:
 
  • #9
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:
 

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

What are complex numbers and why do we need to use them?

Complex numbers are numbers that have both a real and imaginary part. They are represented in the form a + bi, where a is the real part and bi is the imaginary part. We need to use complex numbers because they allow us to solve certain mathematical problems that cannot be solved using only real numbers. They are also used in various fields of science and engineering, such as electrical engineering and quantum mechanics.

How do you add and subtract complex numbers?

To add or subtract complex numbers, we simply add or subtract the real and imaginary parts separately. For example, to add (3 + 2i) and (1 + 4i), we add 3 + 1 for the real parts and 2i + 4i for the imaginary parts, giving us the result 4 + 6i.

Can you multiply and divide complex numbers?

Yes, complex numbers can be multiplied and divided just like real numbers. To multiply, we use the distributive property and simplify using the fact that i^2 = -1. To divide, we multiply by the complex conjugate of the denominator and simplify. The result will be a complex number in the form a + bi.

What is the complex conjugate of a complex number?

The complex conjugate of a complex number a + bi is the number a - bi. In other words, the complex conjugate has the same real part but the imaginary part has the opposite sign. This is important when dividing complex numbers, as it allows us to get rid of the imaginary part in the denominator.

How are complex numbers used in engineering and physics?

Complex numbers have many applications in engineering and physics. In electrical engineering, they are used to represent AC circuits and signals. In physics, they are used to describe quantum mechanical systems and electromagnetic waves. They also have applications in fluid dynamics, control systems, and signal processing.

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