Question

forma trigonometrica a nr complex 2i este ? Cu explicatie, va rog!

Answer 1

$\displaystyle\bf\\fie~z=a+bi~\rightarrow~(a,b),~atunci~forma~trigonometrica~a~acestui~\\numar~este~data~de~:~z=r(cos~\alpha+i\cdot sin~\alpha),~r=modulul~numarului~\\complex,~r=\sqrt{a^2+b^2},~iar~\alpha~argumentul~numarului~complex,~\\\alpha\in[0,2\pi],~\alpha~se~obtine~din~ecuatia~:~tg~\alpha = \frac{b}{a},~\implies\\\alpha_k=arctg\frac{b}{a} +k\pi,~din~care~vom~alege~unghiul~\alpha~ce~apartine~cadranului~\\corespunzator~imaginii~geometrice.\\in~cazul~nostru~:\\z=2i=0+2i \rightarrow (0,2).$

$\displaystyle\bf\\primul~pas~este~sa~reprezentam~intr-un~sistem~de~axe~imaginea~\\geometrica~a~numarului~complex~(vezi~in~atasament).\\r=\sqrt{0^2+2^2} = 2,~iar~\alpha ~=\frac{\pi }{2},~asadar~:~z=r(cos~\alpha~+i\cdot sin~\alpha ~) = \\2(cos\frac{\pi }{2} + i\cdot sin\frac{\pi }{2} )$