Easy complex # question (1 Viewer)

[Giant Lobster]

Yes, it is easy, and i cannot do it cos i suck

Anyway, here it is:

On an Argand diagram the points P and Q represent the numbers z_1 and z_2 respectively. OPQ is an equilateral triangle. SHow that (z_1)^2 + (z_2)^2 = (z_1)(z_2)

thanks

 

[underthesun]

I'm stuck too:

let p = z_1, q = z_2

because OPQ is equilateral, q = p(cis(pi/3))

Trying direct substitution,

p^2 + q^2 = pq

p^2 + (p^2)cis(2pi/3) = (p^2)cis(pi/3)

divide all by p^2,

1 - cis(pi/3) + cis(2pi/3) = 0

1 - cos(pi/3) - isin(pi/3) + (cos(2pi/3) = -cos(pi/3)) + (isin(2pi/3) = isin(pi/3)) = 0

Now you can cancel the imaginary part.

1 - cos(pi/3) - cos(pi/3) = 0

and woah, cos(pi/3) turns out to be 0.5

1 - 0.5 - 0.5 = 0

wow, i worked it out. I thought i was stuck too..

 

[freaking_out]

hey so can u actually work backwards like that??

 

[MyLuv]

OPQ is equilateral--> z2=z1*cis60 (suppose z2 is on RHS of z1)-->z2^2=z1^2* (cis60)^2=z1^2*cis120
z1*z2= (z1)^2 * cis60
z1^2 +z2^2= z1^2 +z1^2*cis120 =z1^2(1+cis120)=z1^2*cis60
---> z1^2+z2^2= z1*z2

 

[MyLuv]

Btw, how come all letters in the forum suddenly get too big???

 

[freaking_out]

Originally posted by MyLuv
Btw, how come all letters in the forum suddenly get too big???

??? maybe, its gotta do with the text size set by ya browser...if u have IE, then try going to view>text size...and change the size, c if that works.

 

[underthesun]

Originally posted by freaking_out
hey so can u actually work backwards like that??

I'm not quite sure, and I dont' think i'll actually do that in an exam.

Anyways, in the half yearly I got full mark for a part that I worked on backwards like that, so there

 

[Affinity]

hmm working backwards:

Idiots are humans
humans include (name)
therefore (name) is an idiot

 

[McLake]

You can always work backwards up the page!

 

[freaking_out]

Originally posted by McLake
You can always work backwards up the page!

na, i don't think that u can do that in this case...

 

[freaking_out]

Originally posted by underthesun
I'm not quite sure, and I dont' think i'll actually do that in an exam.

Anyways, in the half yearly I got full mark for a part that I worked on backwards like that, so there

yeah, i asked coz my teacher said that u can't do that coz u're assuming the statement to b correct in the first place.

 

[underthesun]

Originally posted by freaking_out
na, i don't think that u can do that in this case...

Let me see, I'll try that but not assuming that the case is true:

0 = 0

1 - 1 = 0

1 - 0.5 - 0.5 = 0

Because cos(pi/3) = 0.5 and cos(2pi/3) = -0.5,

1 - cos(pi/3) + cos(2pi/3) = 0 ....... equation (blah)

now:
sin(pi/3) = sin(2pi/3)

isin(pi/3) = isin(2pi/3)

isin(2pi/3) - isin(pi/3) = 0

adding the above to equation blah,

1 - cos(pi/3) - isin(pi/3) + cos(2pi/3) + isin(2pi/3) = 0

1 - cis(pi/3) + cis(2pi/3) = 0

1 + cis(2pi/3) = cis(pi/3)

multiply both side by p^2

p^2 + p^2*cis(2pi/3) = p*p(cis(pi/3))

let q = pcis(pi/3)

p^2 + (p(cis(pi/3)))^2 = pq

p^2 + q^2 = pq

now, p = z_1, q = z_2

z_1^2 + z_2^2 = (z_1)(z_2)

Basically what I wrote previously, having no idea where I was going, but this time I wrote it backwards up the page.

Alternatively, I wonder if you can write in the exam : "Please read from the end of the solution backwards, bleh bleh blah" etc..

 

[freaking_out]

Originally posted by underthesun
Let me see, I'll try that but not assuming that the case is true:

0 = 0

1 - 1 = 0

1 - 0.5 - 0.5 = 0

lol, it looks so funny when it is that way....

 

[redslert]

i was told that you can work backwards.....all you have to do is write at the beginning, "assume statement is true"

it's just like doin something by just looking at it, you write "by inspection" so that the marker knows that you did it in your head