Textbook Mistakes (Complex)

1. Oct 7, 2004

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?

2. Oct 7, 2004

Tide

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

3. Oct 8, 2004

HallsofIvy

Staff Emeritus
"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"?

4. Oct 8, 2004

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?

5. Oct 8, 2004

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.

6. Oct 8, 2004

HallsofIvy

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

7. Oct 8, 2004

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?

8. Oct 9, 2004

JasonRox

So I guess it's ok.

9. Oct 9, 2004

HallsofIvy

Staff Emeritus
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.

10. Oct 9, 2004

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.

Last edited: Oct 9, 2004
11. Oct 9, 2004

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.

12. Oct 9, 2004

Muzza

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

13. Oct 9, 2004

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.

Last edited: Oct 9, 2004
14. Oct 10, 2004

Chronos

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

15. Oct 10, 2004

JasonRox

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

16. Oct 11, 2004

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.

17. Oct 11, 2004

HallsofIvy

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

In particular, the modulus of (1- t2)+ 2t i is
$$\sqrt{(1-t^2)^2+ 4t^2}= \sqrt{1- 2t^2+ t^4+ 4t^2}$$
$$= \sqrt{1+ 2t^2+ t^4}= \sqrt{(1+ t^2)^2}= 1+t^2. Since the denominaotr of $\frac{(1-t^2)+ 2ti}{1+t^2}$ IS the positive real number 1+t2, the modulus is 1. (That's assuming, of course, that t is a real number, which, as matt grime pointed out, you never did say.) 18. Oct 11, 2004 JasonRox Oh, I wasn't sure you can find the modulus of the numerator on its own. (I believe that is what you are doing.) Yes, t is real. I'll be more informative next time. 19. Oct 12, 2004 matt grime [tex]\frac{a+ib}{c}=\frac{a}{c}+ \frac{ib}{c}$$

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.

20. Oct 12, 2004

JasonRox

I get it now even better now!

Thanks, guys.