Question

Calcule (1+j)^35, De a resposta na forma algébrica.​

Answer 1

z=1+i

ρ²=1²+1²

ρ=√2

tgθ=1/1

θ=π/4

${z}^{n} = {ρ}^{n} ( \cos(nθ) + isen(nθ))$

${(1 + i)}^{35} \\ = {( \sqrt{2} ( \cos( \frac{\pi}{4} )+ i \sin( \frac{\pi}{4}) )}^{35}$

${ \sqrt{2} }^{35} ( \cos( \frac{35\pi}{4}) + i \sin( \frac{35\pi}{4} ) ) \\ = { \sqrt{2} }^{35} .( - \frac{ \sqrt{2} }{2} + i \frac{ \sqrt{2} }{2} )$