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Modulul numărului complex z=(1+i)^10 este..?
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Răspunsuri la întrebare
z∣=∣2⋅(1−i)
z∣=∣2⋅(1−i) 3
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣=
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1)
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 =
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2(
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 )
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 ) 3
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 ) 3 =4
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 ) 3 =4 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 ) 3 =4 2
z∣=∣2⋅(1−i) 3 ∣=2∣1−i∣ 3 .Dar |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}∣1−i∣= 1 2 +(−1) 2 = 2 .Rezulta ca |z| = 2 (\sqrt{2})^3 = 4 \sqrt{2}.∣z∣=2( 2 ) 3 =4 2 .
Răspuns:
Sper ca te-am ajutat...Bafta
