# QUICK HELP complex numbers

1. Nov 23, 2004

### aisha

Let Z1 = 3-i
Z2=7+2i express (1/Z1)-(1/Z2) in form a+bi
SOMEONE plzzzzzzz HELP ME!!! I dont have a clue as to how to do this
What do I do?
Where do I start?

2. Nov 23, 2004

### 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! :)

3. Nov 23, 2004

### Parth Dave

Do you know what a "complex conjugate" is?

4. Nov 23, 2004

### aisha

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 dont know what to do or what order to do it in plz help me plzzz

5. Nov 23, 2004

### BobG

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.

6. Nov 23, 2004

### aisha

Ive never rationalized the denominator of a fraction so im 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 dont get the right answer plz show me how do i divide first subtract use conjugates or what?

7. Nov 23, 2004

### 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?

8. Nov 23, 2004

### aisha

a)89/24-73/24i

b)89/24+73/24i

c)-89/24+73/24i

d)89/24i-73/24

I dont get anything close to these when i take those two answers and subtract them. Please help me Ive been doing this one forever

9. Nov 24, 2004

### BobG

All of their choices are wrong.

So what did you get? Something close to

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

10. Nov 24, 2004

### 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)

11. Nov 24, 2004

### arildno

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

12. Nov 24, 2004

### BobG

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!"

13. Nov 24, 2004