Man har komplexa talen

Har jag gjort rätt?

$u \cdot z = r_{u} \cdot r_{z} (\cos (v_{u} + v_{z}) + i \sin (v_{u} + v_{z})) = 2 \cdot 6 (\cos (- \frac{5 π}{4} + \frac{5 π}{3}) + i \sin (- \frac{5 π}{4} + \frac{5 π}{3})) = 12 (\cos (- \frac{15 π + 20 π}{12}) + i \sin (- \frac{15 π + 20 π}{12})) = 12 (\cos (\frac{5 π}{12}) + i \sin (\frac{5 π}{12})) w \cdot w = r_{w} \cdot r_{w} (\cos (v_{w} + v_{w}) + i \sin (v_{w} + v_{w})) = 4 \cdot 4 (\cos (\frac{π}{3} + \frac{π}{3}) + i \sin (\frac{π}{3} + \frac{π}{3})) = 16 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})) \frac{u \cdot z}{w^{2}} = \frac{r_{u z}}{r_{w^{2}}} (\cos (v_{u z} - v_{w^{2}}) + i \sin (v_{u z} - v_{w^{2}})) = \frac{12}{16} (\cos (\frac{5 π}{12} - \frac{2 π}{3}) + i \sin (\frac{5 π}{12} - \frac{2 π}{3}) = \frac{3}{4} (\cos (\frac{5 π - 8 π}{12}) + i \sin (\frac{5 π - 8 π}{12})) = \frac{3}{4} (\cos (- \frac{3 π}{12}) + i \sin (- \frac{3 π}{12})) = \frac{3}{4} \cos (- \frac{π}{4}) + i \sin (- \frac{π}{4}))$

ges i intervallet mellan 0 och 2pi

$- \frac{π}{4} + n \cdot 2 π = - \frac{π}{4} + 2 π = - \frac{π}{4} + \frac{8 π}{4} = \frac{7 π}{4} = \frac{3}{4} (\cos \frac{7 π}{4} + isin \frac{7 π}{4})$

b) uppgiften

ge exakta svaret på formen a+bi

$\frac{3}{4} (\cos \frac{7 π}{4} + isin \frac{7 π}{4}) = \frac{3}{4} ((- \frac{1}{\sqrt{2}}) + i (- \frac{1}{\sqrt{2}})) = - \frac{3}{4 \sqrt{2}} - \frac{3}{4 \sqrt{2}} i = - \frac{3 \sqrt{2}}{4 \sqrt{2} \sqrt{2}} - \frac{3 \sqrt{2}}{4 \sqrt{2} \sqrt{2}} i = - \frac{3 \sqrt{2}}{4 \sqrt{4}} - \frac{3 \sqrt{2}}{4 \sqrt{4}} i = - \frac{3 \sqrt{2}}{4 \cdot 2} - \frac{3 \sqrt{2}}{4 \cdot 2} i = \frac{3 \sqrt{2}}{8} - \frac{3 \sqrt{2}}{8} i$

Är detta korrekt?

Jag hittar inga fel i det.

Bubo skrev:
Jag hittar inga fel i det.

Yay! Tack!

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