AmTran_Bus
  • AmTran_Bus
Really confused all I need is explanation and maybe some steps.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AmTran_Bus
  • AmTran_Bus
Express 6i in the form re^(i theta).
AmTran_Bus
  • AmTran_Bus
@ParthKohli @paki @Nnesha
AmTran_Bus
  • AmTran_Bus
I have the steps. My book says 6i = r cos theta + ir sin theta Then r=6. How?

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AmTran_Bus
  • AmTran_Bus
I understand why we use Euler's formula.
ParthKohli
  • 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|
ParthKohli
  • ParthKohli
|dw:1440264204510:dw|
AmTran_Bus
  • AmTran_Bus
That is what @phi and @IrishBoy123 said, too, but I still need to understand the actual math.
AmTran_Bus
  • AmTran_Bus
Like, I see and think I get the graph. My book has it.
phi
  • 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
phi
  • phi
thus if you start with 0 + 6 i and find the magnitude, you will get 6 that means r = 6
AmTran_Bus
  • AmTran_Bus
How did you get the 0 in 0+6i?
phi
  • phi
6i is 0 + 6i (if we insist on showing the real component is 0)
AmTran_Bus
  • AmTran_Bus
Oh. The z=x+iy formula. Ok.
AmTran_Bus
  • AmTran_Bus
So now I need tan-1, which is 6/0?
ParthKohli
  • ParthKohli
This is the actual math. If you want to look at it the boring way, here you go:\[\tan \theta=\frac{y}{x} = \frac{\Im(z)}{\Re(z) } = \frac{6}{0}\]\[\Rightarrow \theta = \pi/2\]\[r = \sqrt{x^2 + y^2} = \sqrt{\left(\Re(z)\right)^2 + \left(\Im(z)\right)^2} = 6\]
phi
  • phi
yes, the angle can be found using inverse tan imag/real
AmTran_Bus
  • AmTran_Bus
Thanks! So to express it in that certain form, It must be 6e^(i pi/2)
phi
  • phi
yes, you are using Euler's \[ e^{ix}= \cos x + i \sin x \]
IrishBoy123
  • IrishBoy123
|dw:1440264786581:dw|
AmTran_Bus
  • AmTran_Bus
Why does my calc say tan 6/0 is undef and yall get pi/2?
IrishBoy123
  • IrishBoy123
because 6/0 is undefined
IrishBoy123
  • IrishBoy123
ask it for \(arctan \frac{6}{0.00000001}\)
AmTran_Bus
  • AmTran_Bus
got 1.57
AmTran_Bus
  • AmTran_Bus
ok tks
IrishBoy123
  • IrishBoy123
my calc blows up too. its software just crashes on a singularity, it doesn't ask which function is coming next....
AmTran_Bus
  • AmTran_Bus
Can yall help me with one more? 4-sqrt(2)i?
AmTran_Bus
  • AmTran_Bus
Do I need to multiply it by 4+sqrt(2)i?
phi
  • phi
you can do sum of squares (then take the square root)
AmTran_Bus
  • AmTran_Bus
Ok. Let me get that ans for ya!
AmTran_Bus
  • AmTran_Bus
Its just 16+2 right?
phi
  • phi
yes if you plot it, you see it's pythagoras
AmTran_Bus
  • AmTran_Bus
Why does the book say sqrt(16+2) and then 3 sqrt(2)?
phi
  • phi
|dw:1440265488511:dw|
AmTran_Bus
  • AmTran_Bus
If I took a picture of the book, would you help me make sense of what its saying?
phi
  • phi
do you get that the length (distance from the origin) of the point 4 + sqr(2) I is sqr(18)?
AmTran_Bus
  • AmTran_Bus
Yes. But this book does not do it that way.
AmTran_Bus
  • AmTran_Bus
It starts under the 6i
phi
  • 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 \[ 3 \sqrt{2}\]
AmTran_Bus
  • 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.
AmTran_Bus
  • AmTran_Bus
oh you take sqrt to get mag.
phi
  • 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.
AmTran_Bus
  • AmTran_Bus
Yea, I see where pythagoras helps understand that now.
AmTran_Bus
  • AmTran_Bus
Thanks for that.
IrishBoy123
  • IrishBoy123
this is from the Gospel of St Mary [Boas, Maths for Physical Sciences] notice how the Argand diagram features prominently
1 Attachment
AmTran_Bus
  • AmTran_Bus
Wow that is really nice to have. Thanks!
AmTran_Bus
  • AmTran_Bus
Just solved 3 on my own got em all right
IrishBoy123
  • IrishBoy123
https://gyazo.com/a85e151f5d058e0a0a631873c5c59849
IrishBoy123
  • IrishBoy123
you have 30 seconds!

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