## Loser66 one year ago Someone explains me, please. Question on attachment.

1. Loser66

2. Loser66

I don't get "Equivalently..." why is it?

3. Loser66

@ganeshie8

4. ganeshie8

Can the magnitude of a complex number ever be negative ? |dw:1440808727196:dw|

5. Loser66

No

6. Loser66

but zw (bar) is not the magnitude.

7. ganeshie8

well the real part of $$z\overline{w}$$ is magnitude of something

8. ganeshie8

|dw:1440809022356:dw|

9. Loser66

Question: when a complex number is consider as a negative one?

10. ganeshie8

complex numbers don't have that sense of binary negative /positive the sense comes from the arguement let me think a bit..

11. ganeshie8

the magnitude of $$z\overline{w}$$ equals the real component of $$z\overline{w}$$ can we conclude that the imaginary component of $$z\overline{w}$$ is 0 ?

12. Loser66

Yes, I think so

13. ganeshie8

since the imaginary component is 0, the complex number $$z\overline{w}$$ is just as good as a real number and we can compare it with other reals like 0 etc

14. ganeshie8

Moreover $$z\overline{w}$$ identically equals the magnitude of something, so it has to be nonnegative

15. Loser66

No, not that, like |dw:1440809469436:dw|

16. ganeshie8

By definition, ,magnitude of a complex number is the distance from itself to the origin. Length can never be negative

17. ganeshie8

|dw:1440809618870:dw|

18. Loser66

I know, I talk about the vector itself, $$z\overline w$$, not $$|z\overline w|$$

19. ganeshie8

$$\large -5+0i = -5$$ is on negative side, so it can never be magnitude of somebody.

20. Loser66

We have $$|z\overline w|=Re (z\overline w)$$ I am ok with that, but Equivalently $$z\overline w$$, (this guy is not the magnitude or a Real part of anything, and it is $$\geq 0$$ That is what I don't know how to argue.

21. ganeshie8

we have concluded earlier that the imaginary part of $$z\overline{w}$$ is 0, yes ?

22. Loser66

Yes

23. ganeshie8

therefore $$\Re{(z\overline{w})} = z\overline{w}$$

24. Loser66

ok, got you. hehehe.. yes. Thanks a lot.

25. ganeshie8

np

26. ganeshie8

the solution in docx kinda sucks... its not detailed and you will need to put together the missing pieces

27. ganeshie8

|dw:1440810132148:dw|

28. Loser66

oh, for this part, I know how to fix.

29. ganeshie8

can you explain me that part is that some identity that you're using or do you know why is it true

30. Loser66

We have |a| = |b| iff |a|^2 =|b|^2 Apply to this , we have: $$||z|-|w||= |z -w|$$ iff $$LHS ^2 = RHS^2$$

31. ganeshie8

got that part

32. Loser66

$$LHS^2 = (|z|-|w|) \overline{(|z|-|w|)} =(|z|-|w|)(|\overline z| -\overline{|w|})$$

33. ganeshie8

for LHS aren't you simply using the identity $(a-b)^2 = a^2-2ab+b^2$ ?

34. Loser66

$$=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2$$

35. Loser66

The middle terms only $$\overline{z\overline{w}}= \overline{z}w$$

36. Loser66

Hence it is $$-(z\overline{w} +\overline{z\overline{w}})= -2Re(z\overline{w})$$

37. Loser66

That is why it becomes as shown.

38. Loser66

|dw:1440810939082:dw|

39. Loser66

I did

40. ganeshie8

$$=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2$$ can you show me how the middle two terms equal to $$2\Re(z\overline{w})$$

41. ganeshie8

i don't get it, let me try reading it again

42. Loser66

|dw:1440811335931:dw|

43. ganeshie8

Got it! thank you :)

44. Loser66

|dw:1440811409420:dw|

45. Loser66

hihihi... I have to say thanks to you :)

46. ganeshie8

looks neat! just writing it again : $|z-w|^2 = (z-w)(\overline{z-w}) =(z-w)(\overline{z}-\overline{w})\\~\\= |z|^2\color{blue}{-w\overline{z} -z\overline{w}}+|w|^2 = |z|^2\color{blue}{-\overline{\overline{w}z}-z\overline{w}}+|w|^2= |z|^2\color{blue}{-2\Re{(\overline{w}z})}+|w|^2$

47. Loser66

Thank you. :)