n00bie complex number q's (1 Viewer)

[Xayma]

z normally stands for a complex number (ie z=x+iy or z=kcis@) in that case it was only when |z|=1 so therefore z=cis@

 

[...]

can i just ask for interest

why is it in the form of cos@ + i.sin@
not
i.cos@ + sin@

does it make a difference?

 

[ND]

It makes a huge difference. (it's equivalent to inverting it)

Look at the diagram. It's because a=cos@ and b=sin@. (just by using the trig ratios)

edit: err forgot to attach diagram. Refer to post below.

 

[ND]

.

 

[...]

ah okok..icic

but that only applies if the imaginary plane is vertical..but what happens if the imaginary plane is vertical and real is vertical..

 

[nike33]

errr

 

[:: ck ::]

u cant have an imaginary plane verticle and a real one vertical too...

coz... u JUST CANT [i cant explain it .. hahaha.. mb its coz its the way the argand diagram is defined as?]

hehe not a good explanation =\ ...

 

[nike33]

think outside the square ryan....haha the possibilities everything would be like a dot or a line

 

[ND]

Imaginary is always vertical, and real is always horizontal.

 

[:: ck ::]

lolz thinking of im(z) and re(z) as parallel gives me a headache... hahaha

 

[ND]

Well if they were both vertical, a+bi would appear as a pt on the y axis, (a+b) units above the origin. The obvious problem would be that you don't know how much is real and how much imaginary.

 

[...]

ok, n00bie complex number q #127851

consider the equation | z + 8 | = | z - 6i |
i)by writing z = x + iy, show that the equation can be written in the form 4x + 3y + 7 = 0

ii)in an argand diagram, indicate the pts representing the complex numbers -8 and 6i as well as the line 4x + 3y + 7 = 0.

lol..thanks..

 

[ND]

Come on man that's just straight bashing... This isn't the answer they want (cos they specifically say "by writing z=x+iy", but i'll leave that method to you), but what it's actually asking is, what is the locus of z so that it is equidistant from the points 8 and 6i. Now i'm sure you remember how to do that from 3u.

 

[riVa0o]

oooh n00bie complex no.s! finally one i *might* be able to post up before anyone else =D

|z+8| = |z-6i|
(x+8)^2 + y^2 = x^2 + (y-6)^2 (just using |z-a-bi| = sqrt[(x-a)^2 + (y-b)^2])
expand and you'll end up with 4x+3y+7 = 0

second part, pt for -8 is (-8,0) pt for 6i is (0,6) and line 4x+3y+7=0 is jus the perpendicular bisector of the line between the two points

EDIT: dammit, knew someone would beat me to it ><

 

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Originally posted by ND
but what it's actually asking is, what is the locus of z so that it is equidistant from the points 8 and 6i. Now i'm sure you remember how to do that from 3u.

ah fuck, now its clicking...
i shouldn't be doing maths
-_____-;;

Originally posted by riVa0o
oooh n00bie complex no.s! finally one i *might* be able to post up before anyone else =D

|z+8| = |z-6i|
(x+8)^2 + y^2 = x^2 + (y-6)^2 (just using |z-a-bi| = sqrt[(x-a)^2 + (y-b)^2])
expand and you'll end up with 4x+3y+7 = 0

second part, pt for -8 is (-8,0) pt for 6i is (0,6) and line 4x+3y+7=0 is jus the perpendicular bisector of the line between the two points

EDIT: dammit, knew someone would beat me to it ><

oooh..i'll look at that when i fuck up

 

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hehehe..heres one more

| z - 1 - i | = | z - 5 +i |

heres my working

z = x + iy

| (x - 1) + (y - 1)i |² = | (x - 5) + (y + 1)i |²

(x - 1)² + (y - 1)² = (x - 5)² + (y +1)²

cont'd

 

[Xayma]

Seems right, just expand and simplify.

x²-2x+1+y²-2y+1=x²-10x+25+y²+2y+1
8x-4y-24=0
2x-y-6=0

 

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hmm..time to revive the old!!

heheh..

compute (sqrt(3)/2 - 1/2i)⁷⁷
Give your answer in rectangular form.

what does it mean by rectangular form

i thought there was only polor form..or is that the question is in polar form already, but we need to change it to rectangular..

hmm..

n00b attack! =p

 

[Xayma]

Umm rectangular form would be in the x+iy form, instead of the polar form.

 

[:: ck ::]

yes to wut xayma said [dont expand that using binomial.. to the power of 77 kinda gives u a clue that ur not meant to expand it lolz...]

convert sqrt 3/2 -1/2 i to polar form using the conversoin

x+iy = rcis@ , where r = x^2 + y^2, @ = tan^-1y/x ...

after that apply demoivres... [rcis@]^n = r^ncis(n@)