Really confused all I need is explanation and maybe some steps.

- AmTran_Bus

Really confused all I need is explanation and maybe some steps.

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- AmTran_Bus

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

- AmTran_Bus

@ParthKohli @paki @Nnesha

- AmTran_Bus

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

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## More answers

- AmTran_Bus

I understand why we use Euler's formula.

- 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

|dw:1440264204510:dw|

- AmTran_Bus

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

- AmTran_Bus

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

- 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

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

- AmTran_Bus

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

- phi

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

- AmTran_Bus

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

- AmTran_Bus

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

- 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

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

- AmTran_Bus

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

- phi

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

- IrishBoy123

|dw:1440264786581:dw|

- AmTran_Bus

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

- IrishBoy123

because 6/0 is undefined

- IrishBoy123

ask it for \(arctan \frac{6}{0.00000001}\)

- AmTran_Bus

got 1.57

- AmTran_Bus

ok tks

- IrishBoy123

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

- AmTran_Bus

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

- AmTran_Bus

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

- phi

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

- AmTran_Bus

Ok. Let me get that ans for ya!

- AmTran_Bus

Its just 16+2 right?

- phi

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

- AmTran_Bus

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

- phi

|dw:1440265488511:dw|

- AmTran_Bus

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

- phi

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

- AmTran_Bus

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

##### 1 Attachment

- AmTran_Bus

It starts under the 6i

- 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

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

oh you take sqrt to get mag.

- 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

Yea, I see where pythagoras helps understand that now.

- AmTran_Bus

Thanks for that.

- 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

Wow that is really nice to have. Thanks!

- AmTran_Bus

Just solved 3 on my own got em all right

- IrishBoy123

https://gyazo.com/a85e151f5d058e0a0a631873c5c59849

- IrishBoy123

you have 30 seconds!

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