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Post by mondo on Sept 25, 2022 22:16:08 GMT
Mondo Here's a piece of advice: Why don't you do the exercises at the end of each chapter? It's the only way to make sure you really understand what you have read. Most people start at chapter 1 and then do chapter 2 and then chapter 3 and so on. You started at chapter 10 or 11 and then chapter 2 and now chapter 3 without doing any exercises. This is not a good idea as your questions demonstrate. Why do you do things like this? Are you in a hurry? Vasco Well as most people I also started at chapter 1, 2,.. and then discovered this forum when I was reading later chapters. So that's why you got an impression that I skipped the first part of the book. Although it is true that in my first read I skipped a lot - and that is why I am going through this material again now. I am definitely not in hurry, however one thing I need to improve is consistency. It happened that I had about a month long breaks from study and that is devastating. So now I try to find some time every day. As for the exercises, I do some, usually one or two at the beginning of exercise section. Although this line that I have difficulty to get here seems to be simple, I don't know why it doesn't make sense to me. Mondo If $(\widetilde{z}-q)=\rho e^{i\phi}$ then $\overline{(z-q)}=(R^2/\rho)e^{-i\phi}$ and so $(\widetilde{z}-q)\overline{(z-q)}=\rho e^{i\phi}(R^2/\rho)e^{-i\phi}=R^2$ $(\widetilde{z}-q)=\rho e^{i\phi}$ ? Did you mean $(z-q)=\rho e^{i\phi}$ ? |
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Post by Admin on Sept 25, 2022 22:33:32 GMT
Mondo
OK I understand the situation now.
Yes you are right sorry, I meant to write
if $(\widetilde{z}-q)=(R^2/\rho)e^{i\phi}$ then $\overline{(z-q)}=\rho e^{-i\phi}$
Vasco
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Post by Admin on Sept 25, 2022 22:56:01 GMT
Mondo
Do you understand it now?
Vasco
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Post by Admin on Sept 26, 2022 6:40:59 GMT
Mondo
If we have 2 complex numbers which point in the same direction and we take the conjugate of one of them and then multiply them together we get a real number.
Complex numbers pointing in the same direction have the same value for $\arg$. When you take the conjugate of a complex number its length stays the same but its $\arg$ changes sign.
So if $a$ and $b$ can be written as $a=Ae^{i\phi}$ and $b=Be^{i\phi}$ then
$a\bar{b}=Ae^{i\phi}Be^{-i\phi}=ABe^{i0}=AB$, which is a real number since $A$ and $B$ are real numbers.
Vasco
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Post by mondo on Sept 26, 2022 7:27:23 GMT
Mondo If we have 2 complex numbers which point in the same direction and we take the conjugate of one of them and then multiply them together we get a real number. Complex numbers pointing in the same direction have the same value for $\arg$. When you take the conjugate of a complex number its length stays the same but its $\arg$ changes sign. So if $a$ and $b$ can be written as $a=Ae^{i\phi}$ and $b=Be^{i\phi}$ then $a\bar{b}=Ae^{i\phi}Be^{-i\phi}=ABe^{i0}=AB$, which is a real number since $A$ and $B$ are real numbers. Vasco Do you understand it now? Yes! Thank you Vasco. I think my problem was I somehow overlooked that quantities like $(\tilde{z} - q)$ are still complex numbers not pure lengths. That is why the conjugate did not make sense to me. Thanks again. |
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