Matematyka.pl

\(\displaystyle{ 2Re ft\{ \frac{2z-1+i}{1+i} =z ^{2}\right\}}\)

\(\displaystyle{ Re ft\{ \frac{z-i}{z+i}=0 \right\}}\)
Z drugiej strony też powinna być klamerka i to po klamerce to jest ułamek, ale niestety nie wiem jak zrobic to poprawnie

Chyba o to chodziło.
Szemek

1)\(\displaystyle{ 2Re( \frac{2z-1+i}{1+i} =z ^{2})}\)

\(\displaystyle{ z=x+iy}\)

\(\displaystyle{ \frac{2(x+iy)-1+i}{1+i}=(x+iy)^2}\)

\(\displaystyle{ \frac{2x+2iy-1+i}{1+i}=x^2+2ixy-y^2}\)

\(\displaystyle{ \frac{(2x+2iy-1+i)(1-i)}{(1+i)(1-i)}=x^2+2ixy-y^2}\)

\(\displaystyle{ \frac{2x-2ix+2iy+2y+2i}{2}=x^2+2ixy-y^2}\)

\(\displaystyle{ x-ix+iy+y+i=x^2+2ixy-y^2}\)

\(\displaystyle{ x-x^2+y^2+y+i(y-x-2xy+1)=0}\)

\(\displaystyle{ Re=x-x^2+y^2+y}\)

\(\displaystyle{ 2Re=2(x-x^2+y^2+y)}\)

2)\(\displaystyle{ Re( \frac{z-i}{z+i}=0)}\)

\(\displaystyle{ z=x+iy}\)

\(\displaystyle{ \frac{x+iy-i}{x+iy+i}=0}\)

\(\displaystyle{ \frac{x+i(y-1)}{x+i(y+1)}=0}\)

\(\displaystyle{ \frac{(x+i(y-1))(x-i(y+1)}{(x+i(y+1))(x-i(y+1))}=0}\)

\(\displaystyle{ \frac{x^2+y^2-1-2ix}{x^2+(y+1)^2}=0}\)

\(\displaystyle{ Re(\frac{x^2+y^2-1-2ix}{x^2+(y+1)^2}=\frac{x^2+y^2-1}{x^2+(y+1)^2}}\)