complex numbers

studos

hello!

in polar form of complex numbers, can the angle be negative? can the angle be >360 ?

thanks!

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Reply 1

TeeEm

Original post by studos

hello!

in polar form of complex numbers, can the angle be negative? can the angle be >360 ?

thanks!

strictly speaking no (by convention)

Reply 2

ETRC

should be from -pi to pi i think
in MEI you do it from -pi > theata >/ pi (>/ means greater than or equal to)

Reply 3

studos

you both disagree

what is the correct?

Reply 4

Manitude

I suppose technically it could be greater than 360° however the convention is to use values between -pi and pi as this covers all eventualities. If you use an angle greater than 2pi or 360° then it can be simplified to a value between -pi and pi.

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Reply 5

TeeEm

Original post by studos

you both disagree

what is the correct?

I do not think we disagree

I say it hardly matters ...
if the calculation is mine I can do as I please (in most cases so long as complex logs are involved) since an argument of 13π/6 or π/6 or -11π/6 is the same number.

However the standard convention for the principal argument is -π < argz <=π

(I passed my exams a long time ago so sometimes I do not stick to conventions).

Reply 6

studos

so r<-10 is the same with r<350 ?

Reply 7

TeeEm

Original post by studos

so θ=-10 is the same with θ=350 ?

of course it is

but look at what you are expected to give as answer in your board/module
in other words follow the conventions of your board.

(edited 9 years ago)

Reply 8

studos

Original post by ETRC

should be from -pi to pi i think
in MEI you do it from -pi > theata >/ pi (>/ means greater than or equal to)

how do you expect someone to understand what you wrote?

what is MEI? haven't you learnt that you need to analyze abbreviations before you use them?

also, what is theata? you mean the greek letter theta? you should also indicate that you name the angle of a complex number as theta, because theta could be anything

Reply 9

studos

can you tell me please:

why we introduce complex number? what problem do they solve, that we couldn't solve with the rest of maths?

I think one application, is to be able to calculate the sum of sinusoidally changing amounts
but why can't we do that by simply adding vectors? and we have to introduce, j=sqrt(-1), to do it?

thanks!

Reply 10

studos

can you tell me please why x<90o (polar form) equals to jx ?

I understand that the 90o degrees, mean that the x vector is on the y axis, which is the imaginary axis

but why this equals j times x? ie. sqrt(-1) times x ?

I mean x can be a real number, like 2, right?

why if that x is on the imaginary number, it becomes sqrt(-1) times 2 ?

Reply 11

username1162972

The reason they use -pi to pi is so that they use
Sin(-x) and cos(-x) rules . In other ranges it may not apply.

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Reply 12

TeeEm

Original post by studos

Who upset you today?

I will try to explain the fellow poster's comments

theata is a place you go watch a play.
M.E.I. stands for M@st@rbation Enhances Intelligence

Happy New Year!

PS is the picure next to the username a picture of yourself?

Reply 13

studos

Original post by TeeEm

PS is the picure next to the username a picture of yourself?

yes, are you racist?

Reply 14

TeeEm

Original post by studos

yes, are you racist?

Not at all as I am very foreign myself.
just interest.

Reply 15

rayquaza17

Original post by TeeEm

TeeEmm, you starting the New Years drinking early??

Original post by studos

Complex numbers let us solve

x^{2}+1=0

Original post by studos

Write an arbitrary complex number as z=a+bj.

z=r(cos\Theta +jsin\Theta )

Where

r=\sqrt{a^{2}+b^{2}}

tan\Theta =\frac{b}{a}

Say our complex number was z=j
We then have zero real part.
In polar form, we would have z=1(cos90+jsin90)

Say our complex number was z=4j
In polar form, we would have z=4(cos90+jsin90)

You have to understand that we measure the angle from the positive x-axis line. So if we move along by 90 degrees, we then have no real-part and have only an imaginary part.

Say our complex number was z=-j
In polar form, we would have z=1(cos(-90)+jsin(-90)) = 1(cos270+jsin270)

As mentioned above, it is conventional to use -90<theta<90
(Or equivalently -pi/2<theta<pi/2 in radians!)

Try using Wolfram Alpha and put in some complex numbers to see what their angles would be: http://www.wolframalpha.com/input/?i=-i
I hope this helps.

Reply 16

TeeEm

Original post by rayquaza17

TeeEmm, you starting the New Years drinking early??

I am drinking right now ...
(I had not started drinking when I typed that comment :smile:

)

(edited 9 years ago)

Reply 17

studos

thanks!

but why we need to solve

. ?

Reply 18

Mr M

Study Forum Helper

Original post by studos

Good grief. Sort your attitude out or you will find the number of potential helpers will diminish.

Reply 19

rayquaza17

Original post by studos

thanks!

but why we need to solve

. ?

http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/

This is pretty much how my lecturer explained it to my class.

BTW: MEI is an exam board, like Edexcel, OCR, AQA. (I am not googling the full names of them to write them out!)