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ok lets stop being amatuer Freud's here how often would music even be a subject of discussion?

\\\frac{\gamma -\beta }{\gamma -\alpha }=cis(\frac{\pi}{3}) \ (*) $ and $\frac{\gamma -\alpha }{\gamma -\beta }=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{\gamma -\beta }{\gamma -\alpha }+\frac{\gamma -\alpha }{\gamma -\beta }=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re...

=$

\\$Using the fact that $|x|^2=x\overline{x}, $ $x$ complex$...

\\a+ib=\frac{(x+i)^2}{2x-i}\\|a+ib|=\left|\frac{(x+i)^2}{2x-i}\right|=\frac{|(x+i)^2|}{|2x-i|}=\frac{|x+i|^2}{|2x-i|}\\a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}

\\LHS\\=z_1^2+z_2^2\\=\left(z_2\times cis(\frac{\pi}{3})\right)^2+z_2^2, $ since $z_1=z_2\times cis(\frac{\pi}{3})\\=z_2^2\times(cis(\frac{2\pi}{3})+1)\\=z_2^2 \times(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}+1)\\=z_2^2\times(\frac{1}{2}+i\frac{\sqrt{3}}{2})\\=z_2 \times z_2\times...

\\z_{1}=z_{2}\times cis(\frac{\pi}{3})\\\frac{z_1}{z_2}=cis(\frac{\pi}{3}) \ (*) $ and $ \frac{z_2}{z_1}=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{z_1}{z_2}+\frac{z_2}{z_1}=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re \left(cis(\frac{\pi}{3})\right)=1\\$Multiply through by...

you're all retarded each to their own, why does it bother you?

yep and since 3 points define a UNIQUE circle, the 4th point of a quad must lie on this circle to be considered cyclic

listen to this

you will find that if the 4th point does not lie on the circumference, then the circle will not (and cannot) exhibit properties of a cyclic quadrilateral

\\z^2=a+ib\\|z^2|=|a+ib|\\|z|^2=|a+ib|\\(\sqrt{x^2+y^2})^2=\sqrt{a^2+b^2}\\x^2+y^2=\sqrt{a^2+b^2} \ (*)\\$Now, $\\(x+iy)^2=a+ib\\(x^2-y^2)+2ixy=a+ib\\$Equating $\Re (z),\\x^2-y^2=a\Rightarrow y^2=x^2-a \ (**)\\$Sub $(**) $ in $(*),\\2x^2=a+\sqrt{a^2+b^2} edit: beaten by the Treb Man

\\\frac{1}{1+z}, $ letting $z=\cos \theta+i \sin \theta\\=\frac{1}{1+\cos \theta +i\sin \theta}\\=\frac{1}{1+(2\cos^2 \frac{\theta}{2}-1)+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\\=\frac{1}{2\cos^2 \frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\\=\frac{1}{2\cos...

\\$Let $z=\cos \theta +i\sin \theta $ and $t=\tan \frac{\theta}{2}:\\\frac{1}{1+z}\\=\frac{1}{1+\cos \theta+i\sin \theta}\\=\frac{1}{1+\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}}\\=\frac{1}{2} \times\frac{1+t^2}{1+it}\\=\frac{1}{2} \times\frac{(1+it)(1-it)}{1+it}\\=\frac{1}{2}...

polynomial division should be in Polynomials it works in exactly the same way as division with real factors e.g. x^3-0x^2+2x+1=0 divide by x-3 divide by x-(1+i) letting x-(1+i)=(x-a) and THEN dividing may be of help to you conceptually

g-g-g-g-g-g-g-geniusssssssssssssssssss

\\Z=\sqrt\frac{1+z}{1-z}\Rightarrow z=\frac{Z^2-1}{Z^2+1}\\$If $z $ lies on the circle $x^2+y^2=1, $ then, $|z|=1\\\therefore |z|=\left|\frac{Z^2-1}{Z^2+1}\right|=1\Rightarrow |Z^2-1|=|Z^2+1|\\$Let...

\\P: \ z=a+ib \ (*)\\Q: \ \frac{1}{z-3}+\frac{17}{3}=X+iY=Z \ (**)\\$From $ (**),\\X=\frac{a-3}{(a-3)^2+b^2}+\frac{17}{3} $ and $ Y=-\frac{b}{(a-3)^2+b^2}\\$But $ |z-3|=3\Rightarrow (a-3)^2+b^2=9 \ (***)\\\therefore X=\frac{a-3}{9}+\frac{17}{3} $ and $ Y=-\frac{b}{9}\\$Rearranging and squaring...

\\Z=\sqrt\frac{1+z}{1-z}\Rightarrow z=\frac{Z^2-1}{Z^2+1}\\$If $z $ lies on the circle $x^2+y^2=1, $ then, $|z|=1\\\therefore |z|=\left|\frac{Z^2-1}{Z^2+1}\right|=1\Rightarrow |Z^2-1|=|Z^2+1|\\ not 100% sure how to proceed

\\Z=\frac{2+z}{2-z}\\z=2 \times \frac{Z-1}{Z+1}\\$Let $Z=x+iy,\\z=2 \times \frac{x+iy-1}{x+iy+1}=2 \times \frac{(x^2+y^2-1)+i(2y)}{(x+1)^2+y^2}\\$If $z $ describes the $y$-axis, then $ \Re (z)=0\\\therefore x^2+y^2-1=0\Rightarrow x^2+y^2=1\\$Hence the point $Z $ describes a circle with radius $1...