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Another way to do it algebraically is to represent the complex numbers as cis(a1) and cis(a2) then use the sum to product trigonometric identities.

yea its all goods imo the whole point of this forum is just to communicate ideas about math and how they can help other people so its not really unexpected u get informal proofs and algebruh errors

the conjugate of \frac{1}{a+bi} is not \frac{a-ib}{a^2+b^2} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2} so the conjugate is \frac{a+bi}{a^2+b^2} so using this method u still get the same answer as i did as u get \frac{(x-iy)(a+bi)}{a^2 + b^2} =\frac{(x-iy)(a+bi)}{(a+bi)(a-bi)} =\frac{x-iy}{a-bi}

the hardest part was literally just typing it on latex after doing it on paper

\frac{(x+iy)(a-ib)}{(a-ib)(a+ib)} =\frac{xa-ixb+iya+yb}{(a-ib)(a+ib)} Taking the conjugate we get =\frac{xa+yb-i(ya-xb)}{(a-ib)(a+ib)} =\frac{a(x-iy) -ib(x-iy)}{(a-ib)(a+ib)} =\frac{x-iy}{a-ib} as required

just consider the conjugate of \frac{x+iy}{a+bi} after realising it its literally not that hard of a problem as u say it is

the question does have a solution ur just being ignorant. equations exist that dont have solutions though like x=x+1 and |x|=-1

next time i cant do a question in a test ima just say it doesnt have a solution

for 15 sketch it out if it helps visualise the situation and use dot product theres quite a few different ways u could do 16 but the most reasonable one imo is to also just sketch it out and get an inequality from that

my physics depth study is worth more than my hsc trials :/

are there any textbooks that are recommended to follow along with the videos for practice questions like how mcgrathematics follows the cambridge textbook for maths?

wikipedias definition of the fundamental theorem of algebra basically says what i said

would this not have 2 complex solutions? 3x^2 +4x+2i = 0 x=\frac{-4 \pm \sqrt{16-24i}}{6} \text{Let } \sqrt{16-24i}=a+bi, \text{where a and b are real numbers} \text{From this we have, } a^2 - b^2 = 16 \text{ and, } ab = -12 \text{Rearranging the second term as } b=\frac{-12}{a}, a^2 -...

they are probably referring to (2k+1)^{2} = 2k^2 +2k +1 which should be (2k+1)^{2} = 4k^2 +4k +1. the proof is still correct otherwise

tf they doin to these reactions to make Keq 1000

doesnt letting z=acis\theta and z=bcis\theta imply that a>0 and b>0 since they are the magnitudes of each complex number

oops yea didnt even notice was too busy trying to figure out latex lol

instead of letting z_{k}=x+iy u could just leave it as z_{k} LHS=|z_{k+1}| =\left|z_{k}\left(1+\frac{i}{|z_{k}|}\right)\right| =\left|z_{k}\left(\frac{|z_{k}|+i}{|z_{k}|}\right)\right| =||z_{k}|+i| Using assumption =|\sqrt{k}+i| Since \sqrt{k} is real and i is imaginary...

also about your z=3+i example, 3+i does not lie on the curve. the cartesian equation would be y=x-1 and (3,1) does not lie on this line.