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Re: HSC 2018 MX2 Marathon how about if re(z) is 2Im(z) and z^2-4i is real I cant get the answer.. the answer is +- (2+i)

Re: HSC 2018 MX2 Marathon thank you Pikachu!!

Re: HSC 2018 MX2 Marathon find z suvh that (Im)z=2 and z^2 is real thank you

could someone go through a step by step on how to get the redox half equations for this CuCl2 + KI --> CuI + I2 + KCl thank you :)

thank you <3

so theres this q, am^2+am+c=0 where m is complex number and a,b,c are real so we need to use conjugate theorems to show that a(m conjugate)^2 +b(m conjugate) +c=0 i thought we cn just say that conjugate of m is a root, but how do we use conjugate theorems :/

Re: HSC 2018 MX2 Marathon yes :)

Re: HSC 2018 MX2 Marathon rove by Induction that z1+z2+zn = z1 +z2 +zn where the LHS has a big conjugate line on the top and RHS has small conjugate lines edit: solved

Re: HSC 2018 MX2 Marathon thank you so much!!

Re: HSC 2018 MX2 Marathon if cube root (x+iy) =X+iY show that 4 (X^2-Y^2) = x/X + y/Y

Hey what if the domain is x between 0 and pi/2 would i say that sinx has no solutions?? both should hold true so..

Re: HSC 2018 MX2 Marathon how do i divide x^3-2-2i by x+1-i

ohhhh thanksss

thank you so much for helping :) it makes sense now :D

how do we know that? by plotting?

heey so suppose I have 5-5i = 5 root 2 (cosx+isinx) now i have to equate the roots to eventually get x is it always the case that I can choose to equate the real or imaginary and it doesn't matter which one? do I always get the same answer??

Re: HSC 2018 MX2 Marathon Ahh thank you!!!!!!!!!!

Re: HSC 2018 MX2 Marathon E+ni is a root for ax2+bx+c=0 where a and b and c are real show that an^2 =aE^2 + bE +c so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/

Re: HSC 2018 MX2 Marathon oh okay thanks :)

Re: HSC 2018 MX2 Marathon find x,y if 2z/(1+i) - 2z/i = 5/(2+i) can't seem to get it,, do I sub x+iy later