Jednačina z na kvadrat = -1 + i√3

TEKST ZADATKA

Odrediti sve kompleksne brojeve $z$ za koje važi:

z^2=-1+i\sqrt3

REŠENJE ZADATKA

z=\sqrt[2]{-1+i\sqrt{3}}

Zapisati broj $-1+i\sqrt3$ u trigonometrijskom obliku $z=|z|\cdot(\cos{(\text{arg}(z))}+i\sin{(\text{arg}(z))}).$

-1+i\sqrt3=2\big(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\big)

Primeniti formulu za korenovanje kompleksnog broja: $\sqrt[n]{|z|} \cdot (\cos{\frac{\text{arg}(z)+2k\pi}{n}}+i\sin{\frac{\text{arg}(z)+2k\pi}{n}}), \quad k\in\{0, 1, ...,n-1\}$

\sqrt{2}\cdot (\cos{\frac{\frac{2\pi}{3}+2k\pi}{2}} + i\sin{\frac{\frac{2\pi}{3}+2k\pi}{2}} ), \quad k \in \{ 0,1 \}

Za $k=0:$

z_0=\sqrt{2}\cdot (\cos{\frac{\frac{2\pi}{3}+2\cdot 0 \cdot \pi}{2}} + i\sin{\frac{\frac{2\pi}{3}+2\cdot 0 \cdot \pi}{2}} ) \\ z_0=\sqrt{2}\cdot (\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}} ) \\ z_0=\sqrt{2} \cdot (\frac{1}{2}+i\frac{\sqrt{3}}{2})\\ z_0=\frac{\sqrt{2}}{2}(1+i\sqrt{3})

Za $k=1:$

z_0=\sqrt{2}\cdot (\cos{\frac{\frac{2\pi}{3}+2\cdot 1 \cdot \pi}{2}} + i\sin{\frac{\frac{2\pi}{3}+2\cdot 1 \cdot \pi}{2}} ) \\ z_0=\sqrt{2}\cdot (\cos{(\pi+\frac{\pi}{3})} + i\sin{(\pi+\frac{\pi}{3})} ) \\ z_0=\sqrt{2} \cdot (-\frac{1}{2}-i\frac{\sqrt{3}}{2})\\ z_0=\frac{\sqrt{2}}{2}(-1-i\sqrt{3})

Rešenja jednačine su:

\frac{\sqrt{2}}{2}(1+i\sqrt{3})\quad \text{i} \quad -\frac{\sqrt{2}}{2}(1+i\sqrt{3})

Započni učenje

Online časovi

$z^2=-1+i\sqrt3$

Beleške

z2=−1+i3z^2=-1+i\sqrt3z2=−1+i3​

$z^2=-1+i\sqrt3$