Complex Numbers

[d0minicz]

i did that but somehow something messed up ...
can someone show me their workings plz

[TrueTears]

Sub in .







equating real with real and imaginary with imaginary.

[d0minicz]

argh shit .......... stupid mistake

thanks alot truetears

[d0minicz]

Hey how do i split into two quadratics ?
thanks

[Flaming_Arrow]

make a substitution u=z^2

[TrueTears]

make a substitution u=z^2

but there is a -2z what do you do then?

[Flaming_Arrow]

make a substitution u=z^2

but there is a -2z what do you do then?

sorry didnt read the quesition prlly

[NE2000]

Hey how do i split into two quadratics ?
thanks

Grouping doesn't get you far in this case I think so try solving over R first. That doesn't work for me. I then try and look for clear complex roots, after a bit of experimenting and mulling over this I got z = 1 + i as a root. CRT, z = 1 - i. Factors: (z - 1 - i)(z - 1 + i). One quadratic: (z^2 - 2z + 2). Then use short division to find the other quadratic.

[TrueTears]

Hey how do i split into two quadratics ?
thanks

Grouping doesn't get you far in this case I think so try solving over R first. That doesn't work for me. I then try and look for clear complex roots, after a bit of experimenting and mulling over this I got z = 1 + i as a root. CRT, z = 1 - i. Factors: (z - 1 - i)(z - 1 + i). One quadratic: (z^2 - 2z + 2). Then use short division to find the other quadratic.

yeah but wouldn't it take ages to guess z = 1 + i as a factor?

[NE2000]

Hey how do i split into two quadratics ?
thanks

Grouping doesn't get you far in this case I think so try solving over R first. That doesn't work for me. I then try and look for clear complex roots, after a bit of experimenting and mulling over this I got z = 1 + i as a root. CRT, z = 1 - i. Factors: (z - 1 - i)(z - 1 + i). One quadratic: (z^2 - 2z + 2). Then use short division to find the other quadratic.

yeah but wouldn't it take ages to guess z = 1 + i as a factor?

yeah it would lol but I think a bit of guesswork (not a very mathematical solution I guess) and staring at the equation may be able to help a bit more. I generally try the solutions 1 + i, 1 + rt(3)i or rt(3) + i in cases where I have no clue what to do

so someone else should probably help d0miniciz with the actual solution.

[NE2000]

Or maybe you could make multiple simultaneous equations. Eg:

(z^2+az+c)(z^2+bz+d)

Where dc = 6
Hence
d=6/c

az^3 + bz^3 = 0
a+b=0

Hence
b=-a

abz^2 +dz^2 + cz^2 = 1
ab + d + c = 1

-a^2 + c/6 + c = 1    [1]

cbz + daz = -2z
cb + da = -2
-ac + ac/6 = -2     [2]

I think I made a mistake somewhere......but anyway, maybe that's a way.

[d0minicz]

lols its alright guys dont worry about it
thanks heaps for ur efforts in the meantime

[d0minicz]

Hi i need help solving this over C:
thanks

[d0minicz]

in the back it says Cis Pi/2 , cis (-5pi/6) and cis (-pi/6) on the argand diagram i am confused.

[TrueTears]

change to polar form yields:

now let



equate yields:

and , where

now and

let



( because must be within )

The rest you can do, just sub them back in and you can change them back to cartesian form.

remember we always take the number 1 less than the power of z for k. So for example if , we would take to . If we would take to