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Irregulaar
@Irregulaar
January 2021
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bonjour je n'arrive pas à montrer cette égalité
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anylor
Bonjour
[1/2 (z+izbarre) - z] / (1+i) × (1 -i) /( 1-i)
on multiplie par la quantité conjuguée
[(1/2 (z+izbarre) - z ) × (1 -i)] / [( 1-i)(1+i) ]
=[(1/2 (z+izbarre) - z ) × (1 -i) ] / 1- i²
= [(-z/2 +izbarre/2) × (1 -i) ]/ 1- (-1)
= [ zbarre/2 - z/2+ ( zbarre +z/2) × i ] / 2
= [ zbarre/2 - z/2+ ( zbarre +z/2) × i ] × 1/2
= [ zbarre - z + ( zbarre +z ) × i ] / 4
pour tout z ∈ C
1 votes
Thanks 1
Irregulaar
merci :)
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[1/2 (z+izbarre) - z] / (1+i) × (1 -i) /( 1-i)
on multiplie par la quantité conjuguée
[(1/2 (z+izbarre) - z ) × (1 -i)] / [( 1-i)(1+i) ]
=[(1/2 (z+izbarre) - z ) × (1 -i) ] / 1- i²
= [(-z/2 +izbarre/2) × (1 -i) ]/ 1- (-1)
= [ zbarre/2 - z/2+ ( zbarre +z/2) × i ] / 2
= [ zbarre/2 - z/2+ ( zbarre +z/2) × i ] × 1/2
= [ zbarre - z + ( zbarre +z ) × i ] / 4
pour tout z ∈ C