Liczby zespolone calineczka17: e^z = e^−z , z=? gdzie z ∊ℂ. wynik z wolframa znam ale czy rozwiązanie jest ok z=a+bi z=|r| [ cos(α) + i*sin(α) ] z= |r| e^iα −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−zatem:−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− L = e^z = e^a+bi = e^ae^bi P = e^−z = e^−a−bi = e^−ae^−bi −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−zestawiamy to co mamy:−−−−−−−−−−−−−−−−−−−−−−−−−−−−− L=P e^ae^bi = e^−ae^−bi −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−myślimy o tym jak o: z= |r| e^iα −−−−−−−−gdzie:−−−−−−−−−−−−−− e^a = |r| oraz e^−a=|r| −−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−z uwagi na to że e^x≥0 dla każdego x ∊ ℛ, więc−−−−−−−−−− −−pytanie kiedy "promienie sa równe"−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e^a=e^−a ========> a=0;==========> e⁰=1−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−zatem, |r|=1−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−zadanie sprowadza się do postaci:−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 1*e^bi = 1*e^−bi =====> e^bi = e^−bi −−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−co interpretujemy jako z= |r| e^iα −−−więc−−− e^αi = e^−αi−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−− −−−−−−widzimy że pachnie sprzężeniem −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−[ już dawno pachniało ]−−−−−−−−−−−−−−−−− więc potraktujemy teraz jako:−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−z=|r| [ cos(α) + i*sin(α) ] cos(α) + i*sin(α) = cos(−α) + i*sin(−α) −−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−co jest równe:−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− cos(α) + i*sin(α) = cos(α) − i*sin(α) −−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−teraz porównujemy cz Re i cz Im−−−−−−−−−−−−−−−−− cos(α) = cos(α) oraz i*sin(α) = −i*sin(α) −−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−zatem widzimy że dla sinus musi się zerować−−−−−−− −−−−−−−−a taka sytuacja ma miejsce tylko dla α = nπ −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−zatem −−−−−−−−−−−−−−−−−−− α = nπ −−−−−−−−−−−przypomnienie że b=α ; a=0 −−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−więc już mamy nasze z = a+bi−−−−−−−−−−−−− &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& z = a+bi −−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−z=0+nπi gdzie n∊ℤ. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& a

jc: o.k. Proponuję wyjść z równania e^2z=1.

Adam0: e^2z=e⁰ 2z=2kπi z=kπi, k∊ℤ