Is My Approach to Simplifying Complex Fractions Correct?

[JasonRox]

I want to make sure I am on the right track here.

Here is the first one:

Find the modulus of:
$$(2-i)^6=|2-i|^6=(\sqrt{5})^6=125$$

Am I doing it right?

I broke it up at first, but then the book says this. The answers are different, or what I got was different.

The second:

$$\frac{1+2ti-t^2}{1+t^2}=1$$

Work the bottom and you can get:

$$(1+t^2)=(t+i)(t-i)=(ti+1)(1-ti)$$

Work the top:

$$(ti+1)^2$$

I can only eliminate one of the two on the bottom.

I went on with division, and then I got to 1.

Is that ok?

 

[Tide]

The first one looks good but what exactly are you trying to do with the second?

 

[HallsofIvy]

"I can only eliminate one of the two on the bottom."

Yes, that leaves the fraction (1-ti)/(1+ ti).

"I went on with division, and then I got to 1.

Is that ok?"

No, it's not! Since the numerator and denominator are different, the fraction CAN'T be 1! HOW did you "go on with division"?

 

[JasonRox]

It's something like,

$$\frac{(ac-bd)+(ad-bc)i}{a^2+b^2}$$.

That's division for complex numbers.

Isn't that right? That came to i, and then the modulus of i is 1? Right?

 

[matt grime]

It's, unclear what you're trying to do with the second one, or at least i though iknew you were solving for t, albeit in some odd way (why not just mutlply by 1+t^2 and solve the resulting quadratic? it's just an equation after all) but I've no idea what the modulus of i being 1 has to do with it.

 

[HallsofIvy]

It's something like,

$$\frac{(ac-bd)+(ad-bc)i}{a^2+b^2}$$.

No, that's wrong
$$\frac{a+bi}{c+di}= \frac{(ac+bd)+(ad-bc)i}{a^2+b^2}$$

Notice that it is "ac+bd", not "ac- bd" as you have.

What you are really doing is multiplying both numerator and denominator by c- di, the "complex conjugate" of c+ di.

The numerator is (a+ bi)(c- di)= ac+ adi- bci- bdi²
= (ac+ bd)+ (ad-bc)i
while the denominator is (c²+ d²).

In particular,
$$\frac{1-ti}{1+ti}= \frac{(1-t^2)- 2it}{1+t^2}$$

 

[JasonRox]

Sorry, honest mistake there.

I did the division that the book showed and it came out to one. Also note that (1-ti) is the denominator.

Note: I am trying to find the modulus, so that's why I got 1 from i. So, lil=1.

This is what it comes out to:

$$\frac{1+ti}{1-ti}=\frac{(1)(1)+(-1)(1)+i[(1)(-1)-(1)(1)]}{1+1}$$

and so it equals -i.

l-il=1, as well.

Is this right?

 

[JasonRox]

So I guess it's ok.

 

[HallsofIvy]

Sorry, honest mistake there.

I did the division that the book showed and it came out to one. Also note that (1-ti) is the denominator.

Note: I am trying to find the modulus, so that's why I got 1 from i. So, lil=1.

This is what it comes out to:

$$\frac{1+ti}{1-ti}=\frac{(1)(1)+(-1)(1)+i[(1)(-1)-(1)(1)]}{1+1}$$

and so it equals -i.

l-il=1, as well.

Is this right?

What happened to the "t"??
$$\frac{1+ti}{1-ti}= \frac{(1+ti)(1+ti)}{(1-ti)(1+ti)}= \frac{(1-t^2)+ 2ti}{1+t^2}$$

That depends upon t and so does its modulus.

You can't just replace "i" by 1 because |i|= 1 . |1+ i| is NOT equal to
1+ 1 just as |1- 3| is NOT 1+ 3.

 

[JasonRox]

Damn. I really messed up.


I got:

$$\frac{1+2it-t^2}{1+t^2}$$

Which is the same, and that's good now.

I still can't find a path that leads to a modulus of one.

Note: I appreciate the help.

 

[geometer]

Jason - I think your method of doing the first one; "find the modulus of (2-i)^6 is incorrect. I think what the problem asks you is to raise the complex number (2-i) to the 6th power and then find the modulus of that complex number - not raise the modulus of (2-i) to the 6th power.

 

[Muzza]

But |(2 - i)^6| is equal to |2 - i|^6... He probably forgot to write out the vertical lines.

 

[JasonRox]

No, vertical lines.

Geometer, that is EXACTLY what I did. See the part where I said I broke it apart, and the answer was different. I'll do it again, but who knows.

 

[Chronos]

What happened to the "t"??
$$\frac{1+ti}{1-ti}= \frac{(1+ti)(1+ti)}{(1-ti)(1+ti)}= \frac{(1-t^2)+ 2ti}{1+t^2}$$

That depends upon t and so does its modulus.

You can't just replace "i" by 1 because |i|= 1 . |1+ i| is NOT equal to
1+ 1 just as |1- 3| is NOT 1+ 3.

I have the same problem. You cannot just drop 't' out of the equation this way.

 

[JasonRox]

I have the same problem. You cannot just drop 't' out of the equation this way.

I understand that, but that doesn't show that the modulus equals one.

 

[matt grime]

Have you even written down what the modulus of

$$\frac{(1-t^2)+ 2ti}{1+t^2}$$

is?

When you posted this question you didn't even state you were attempting to show it had modulus 1. We aren't psychic! Nor did you state t was a real parameter. In fact you just had an equation with no hint as to what we were supposed to make of it.

 

[HallsofIvy]

I understand that, but that doesn't show that the modulus equals one.

The moduls (also "absolute value") of a+ bi is
$$\sqrt{a^2+ b^2}$$.

In particular, the modulus of (1- t²)+ 2t i is
$$\sqrt{(1-t^2)^2+ 4t^2}= \sqrt{1- 2t^2+ t^4+ 4t^2}$$
[tex]= \sqrt{1+ 2t^2+ t^4}= \sqrt{(1+ t^2)^2}= 1+t^2.<br /> <br /> Since the denominaotr of $\frac{(1-t^2)+ 2ti}{1+t^2}$ IS the positive real number 1+t², the modulus is 1.<br /> <br /> (That's assuming, of course, that t is a real number, which, as matt grime pointed out, you never did say.)[/tex]

 

[JasonRox]

Oh, I wasn't sure you can find the modulus of the numerator on its own. (I believe that is what you are doing.)

Yes, t is real.

I'll be more informative next time.

 

[matt grime]

$$\frac{a+ib}{c}=\frac{a}{c}+ \frac{ib}{c}$$

simple algebra.

now find the modulus:

$$\sqrt{\frac{a^2}{c^2}+\frac{b^2}{c^2}}$$

pull out the 1/c^2 and what do you have?

if you'd just worked through it and followed your nose you'd've got the right answer.

 

[JasonRox]

I get it now even better now!

Thanks, guys.