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June 26th, 2012, 05:20 AM   #1
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Complex complex numbers

Hi, complex numbers again,


1.) Where 1 and 2 are subscripts, Another way of proving (z1/z2)* = z1*/z2* is to start with (z1/z2)* x z2*. Show how this can be done.

By letting z1 be a+bi and z2 be c+di, and then expanding LHS and factorizing again, I got RHS therefore proving (z1/z2)* = z1*/z2*
My problem comes with the second part, when I multiply (z1/z2)* by z2*,
z1*/z2* x z2*
Canceling common factors,
Isn't it just z1*?!
How do I prove that (z1/z2)* x z2*= z1*/z2* ?

2.) Prove that for all complex numbers z and w, zw*- z*w is purely imaginary or zero.
By letting z be a+bi and w be c+di and expanding, I got the imaginary part,
which was 2(bc-ad)i. How do I prove that its zero..Is it because it lies on the imaginary y-axis, which includes 0?

Greatly appreciated, Thank you so much!
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June 26th, 2012, 10:43 AM   #2
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Re: Complex complex numbers

Hello Tutu,

For your own convenience, please try to solve both the problems using polar representation of the complex numbers.

(1) Try to show that (z1/z2)* x z2* = z1*
(2) If you get the result 2(bc-ad)i, where a, b, c, and d are reals, your result must be either purely imaginary or zero.

Let me know if you could follow the argument.

Best regards,
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June 26th, 2012, 11:36 AM   #3
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Re: Complex complex numbers

Hi thanks a lot!

For question 1.) I got z1*, but to me the question seems to be asking for me to prove z1*/z2* from (z1/z2)* x z2*. That was how I understood the question, which I think is wrong because (z1/z2)* x z2* gets me z1* and not z1*/z2* .

For question 2.) I understand that with 2(bc-ad)i as a result, my result is purely imaginary because of the i, but I dont see how it equals to 0..

Thank you so much!
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June 26th, 2012, 12:00 PM   #4
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Re: Complex complex numbers

Let me try to clarify:

1. If you can prove that (z1/z2)* x z2* = z1*, then it is equivalent to prove that (z1/z2)* = z1*/z2* provided z2* is not equal to 0.
Hence if you could prove that (z1/z2)* x z2* = z1*, you are done!

2. The result can't be purely imaginary AND 0 at the same time. That is why you have been asked to prove that the result is either purely imaginary OR zero.

Have you understood the argument now?

Best regards,
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June 26th, 2012, 12:08 PM   #5
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Re: Complex complex numbers

Ahh yes. Can I ask, is 0 real or imaginary?
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June 26th, 2012, 12:23 PM   #6
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Re: Complex complex numbers

Quote:
... is 0 real or imaginary?
This question is not so easy to answer. Please see http://www.physicsforums.com/showthread.php?t=206108 for more detail.

However, in the context of your problem, if ad = bc, then you get a zero and otherwise you get a pure imaginary number.

Best regards,
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June 26th, 2012, 01:07 PM   #7
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Re: Complex complex numbers

Thanks!

So 2(bc-ad)i is my result, it is a purely imaginary number. Can I say that since ad does not equal to bc, it is a purely imaginary number?
Actually, which part tells that ad does not equal to bc?

Sorry and thank you!
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June 26th, 2012, 01:56 PM   #8
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Re: Complex complex numbers

Quote:
Originally Posted by Tutu
Thanks!

So 2(bc-ad)i is my result, it is a purely imaginary number. Can I say that since ad does not equal to bc, it is a purely imaginary number?
Actually, which part tells that ad does not equal to bc?

Sorry and thank you!
a,b,c,d could be any real numbers. So bc-ad may or may not be zero.

If bc-ad is non-zero then your result is purely imaginary.

If bc-ad is zero, then your result is zero.

So, your result is purely imaginary, or it's zero.
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June 26th, 2012, 02:07 PM   #9
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Re: Complex complex numbers

I get it, thanks!
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June 26th, 2012, 02:15 PM   #10
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Re: Complex complex numbers

Conventionally, do I have to write all that as a conclusion to the question, just o prove that it's zero OR imaginary?


a,b,c,d could be any real numbers. So bc-ad may or may not be zero.

If bc-ad is non-zero then your result is purely imaginary.

If bc-ad is zero, then your result is zero.

So, your result is purely imaginary, or it's zero.[/quote]
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