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Dewar
@Dewar
January 2021
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Bonjour, j'ai un exercice sur la forme exponentielle du nombre complexe à faire pour demain et je galère. Merci pour votre aide !
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ProfdeMaths1
Verified answer
Z1.z2=exp(i.π/3).exp(-iπ/4)
=exp(i(π/3-π/4))
=
exp(i.π/12)
z1/z2=exp(i.π/3).exp(iπ/4)
=exp(i(π/3+π/4))
=
exp(i.7π/12)
z1^3=(exp(i.π/3))^3
=exp(3i.π/3)
=
exp(i.π)
z1.z2.z3=√2.exp(i.π/3).exp(-iπ/4).exp(i.2π/3)
=exp(i(π/3-π/4+2π/3))
=
exp(i.3π/4)
z3^4=(√2.exp(i.2π/3))^4
=4.exp(i.8π/3)
=
4.exp(i.2π/3)
z2/z3=√2/2.exp(-i.π/4).exp(-i2π/3)
=√2/2.exp(-i(π/4+2π/3))
=
√2/2
.
exp(-i.11π/12)
2 votes
Thanks 1
Dewar
Merci beaucoup !
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Verified answer
Z1.z2=exp(i.π/3).exp(-iπ/4)=exp(i(π/3-π/4))
=exp(i.π/12)
z1/z2=exp(i.π/3).exp(iπ/4)
=exp(i(π/3+π/4))
=exp(i.7π/12)
z1^3=(exp(i.π/3))^3
=exp(3i.π/3)
=exp(i.π)
z1.z2.z3=√2.exp(i.π/3).exp(-iπ/4).exp(i.2π/3)
=exp(i(π/3-π/4+2π/3))
=exp(i.3π/4)
z3^4=(√2.exp(i.2π/3))^4
=4.exp(i.8π/3)
=4.exp(i.2π/3)
z2/z3=√2/2.exp(-i.π/4).exp(-i2π/3)
=√2/2.exp(-i(π/4+2π/3))
=√2/2.exp(-i.11π/12)