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Trigonometrijski i eksponencijalni oblik

TEKST ZADATKA

Odrediti sve kompleksne brojeve zz za koje važi:

z2=−1+i3z^2=-1+i\sqrt3
REŠENJE ZADATKA
z=−1+i32z=\sqrt[2]{-1+i\sqrt{3}}

Zapisati broj −1+i3-1+i\sqrt3 u trigonometrijskom obliku z=∣z∣⋅(cos⁡(arg(z))+isin⁡(arg(z))).z=|z|\cdot(\cos{(\text{arg}(z))}+i\sin{(\text{arg}(z))}).

−1+i3=2(cos⁡2π3+isin⁡2π3)-1+i\sqrt3=2\big(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\big)
DODATNO OBJAŠNJENJE

Primeniti formulu za korenovanje kompleksnog broja: ∣z∣n⋅(cos⁡arg(z)+2kπn+isin⁡arg(z)+2kπn),k∈{0,1,...,n−1}\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\}

2⋅(cos⁡2π3+2kπ2+isin⁡2π3+2kπ2),k∈{0,1}\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:k=0:

z0=2⋅(cos⁡2π3+2⋅0⋅π2+isin⁡2π3+2⋅0⋅π2)z0=2⋅(cos⁡π3+isin⁡π3)z0=2⋅(12+i32)z0=22(1+i3)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:k=1:

z0=2⋅(cos⁡2π3+2⋅1⋅π2+isin⁡2π3+2⋅1⋅π2)z0=2⋅(cos⁡(π+π3)+isin⁡(π+π3))z0=2⋅(−12−i32)z0=22(−1−i3)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:

22(1+i3)i−22(1+i3)\frac{\sqrt{2}}{2}(1+i\sqrt{3})\quad \text{i} \quad -\frac{\sqrt{2}}{2}(1+i\sqrt{3})

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