Bonjour, Pourriez vous m'aidez pour mettre sous forme exponentielle ce nombre complexe: z=(V2+iV2)^3(1-iV3)^5 Les V sont des racines carrées Merci d'avance
D'où z = (2((√2/2)+i(√2/2)))³(2((1/2)-i(√3/2)))⁵ z = (2(cos(π/4)+isin(π/4)))³(2(cos(-π/3)+isin(-π/3)))⁵ z = 2³(cos(π/4)+isin(π/4))³2⁵(cos(-π/3)+isin(-π/3))⁵ z = 2³exp(iπ/4)³2⁵exp(-iπ/3)⁵ z = 2⁸exp(iπ/4)³exp(-iπ/3)⁵ z = 2⁸exp(i3π/4)exp(-i5π/3) z = 2⁸exp(i3π/4)exp(iπ/3) z = 2⁸exp((i3π/4)+(iπ/3)) z = 2⁸exp((i9π/12)+(i4π/12)) z = 2⁸exp((i9π+i4π)/12) z = 2⁸exp(i13π/12) z = 2⁸exp(-i11π/12)
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Bonsoir,Soit z = (√2+i√2)³(1-i√3)⁵ ∈ℂ
D'où z = (2((√2/2)+i(√2/2)))³(2((1/2)-i(√3/2)))⁵
z = (2(cos(π/4)+isin(π/4)))³(2(cos(-π/3)+isin(-π/3)))⁵
z = 2³(cos(π/4)+isin(π/4))³2⁵(cos(-π/3)+isin(-π/3))⁵
z = 2³exp(iπ/4)³2⁵exp(-iπ/3)⁵
z = 2⁸exp(iπ/4)³exp(-iπ/3)⁵
z = 2⁸exp(i3π/4)exp(-i5π/3)
z = 2⁸exp(i3π/4)exp(iπ/3)
z = 2⁸exp((i3π/4)+(iπ/3))
z = 2⁸exp((i9π/12)+(i4π/12))
z = 2⁸exp((i9π+i4π)/12)
z = 2⁸exp(i13π/12)
z = 2⁸exp(-i11π/12)