## AmTran_Bus one year ago Really confused all I need is explanation and maybe some steps.

1. AmTran_Bus

Express 6i in the form re^(i theta).

2. AmTran_Bus

@ParthKohli @paki @Nnesha

3. AmTran_Bus

I have the steps. My book says 6i = r cos theta + ir sin theta Then r=6. How?

4. AmTran_Bus

I understand why we use Euler's formula.

5. ParthKohli

Yeah well, each complex number is an addition of its "real"-part and its "imaginary"-part. Whenever you compare two complex numbers, you compare their real and imaginary parts separately. You know what the best way to find out the polar-form of a complex number is? Graphing it. |dw:1440264084863:dw|

6. ParthKohli

|dw:1440264204510:dw|

7. AmTran_Bus

That is what @phi and @IrishBoy123 said, too, but I still need to understand the actual math.

8. AmTran_Bus

Like, I see and think I get the graph. My book has it.

9. phi

find the magnitude by multiplying by the complex conjugate: $(r \cos x + i r \sin x ) (r \cos x- i r \sin x )= r^2 \cos^2 x + r^2 \sin^2 x \\ = r^2 ( \cos^2 x + \sin^2 x) \\= r^2$ and the magnitude is r

10. phi

thus if you start with 0 + 6 i and find the magnitude, you will get 6 that means r = 6

11. AmTran_Bus

How did you get the 0 in 0+6i?

12. phi

6i is 0 + 6i (if we insist on showing the real component is 0)

13. AmTran_Bus

Oh. The z=x+iy formula. Ok.

14. AmTran_Bus

So now I need tan-1, which is 6/0?

15. ParthKohli

This is the actual math. If you want to look at it the boring way, here you go:$\tan \theta=\frac{y}{x} = \frac{\Im(z)}{\Re(z) } = \frac{6}{0}$$\Rightarrow \theta = \pi/2$$r = \sqrt{x^2 + y^2} = \sqrt{\left(\Re(z)\right)^2 + \left(\Im(z)\right)^2} = 6$

16. phi

yes, the angle can be found using inverse tan imag/real

17. AmTran_Bus

Thanks! So to express it in that certain form, It must be 6e^(i pi/2)

18. phi

yes, you are using Euler's $e^{ix}= \cos x + i \sin x$

19. IrishBoy123

|dw:1440264786581:dw|

20. AmTran_Bus

Why does my calc say tan 6/0 is undef and yall get pi/2?

21. IrishBoy123

because 6/0 is undefined

22. IrishBoy123

ask it for $$arctan \frac{6}{0.00000001}$$

23. AmTran_Bus

got 1.57

24. AmTran_Bus

ok tks

25. IrishBoy123

my calc blows up too. its software just crashes on a singularity, it doesn't ask which function is coming next....

26. AmTran_Bus

Can yall help me with one more? 4-sqrt(2)i?

27. AmTran_Bus

Do I need to multiply it by 4+sqrt(2)i?

28. phi

you can do sum of squares (then take the square root)

29. AmTran_Bus

Ok. Let me get that ans for ya!

30. AmTran_Bus

Its just 16+2 right?

31. phi

yes if you plot it, you see it's pythagoras

32. AmTran_Bus

Why does the book say sqrt(16+2) and then 3 sqrt(2)?

33. phi

|dw:1440265488511:dw|

34. AmTran_Bus

If I took a picture of the book, would you help me make sense of what its saying?

35. phi

do you get that the length (distance from the origin) of the point 4 + sqr(2) I is sqr(18)?

36. AmTran_Bus

Yes. But this book does not do it that way.

37. AmTran_Bus

It starts under the 6i

38. phi

they are doing the same thing, but left out the simplifying step. sqrt(16+2) = sqr(18)= sqr(2*9)= sqr(2)*sqr(9) = sqr(2)*3 or $3 \sqrt{2}$

39. AmTran_Bus

Got ya. Thanks. Its just hard for me to see sqrt(16+2), because I thought 4- sqrt(2)i was not under a root.

40. AmTran_Bus

oh you take sqrt to get mag.

41. phi

if you think pythagoras with sides 4 (along the x-axis ) and -sqr(2) (along the y or "i" axis) and c^2 = a^2 + b^2 that should help you remember that x^2 + y^2 is the magnitude squared thus 16+2 = 18 is the mag squared and we want sqr(18) for the actual magnitude.

42. AmTran_Bus

Yea, I see where pythagoras helps understand that now.

43. AmTran_Bus

Thanks for that.

44. IrishBoy123

this is from the Gospel of St Mary [Boas, Maths for Physical Sciences] notice how the Argand diagram features prominently

45. AmTran_Bus

Wow that is really nice to have. Thanks!

46. AmTran_Bus

Just solved 3 on my own got em all right

47. IrishBoy123
48. IrishBoy123

you have 30 seconds!