Números complexos
Sendo Z= 1+3i calcule z^2
(Explicando por favor!!)
michelyuna:
No livro tem 12+2.1.3i+(3i^2) e o resto, só não intendo de onde veio essa parte do 12
Soluções para a tarefa
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Vamos lá:
![z = 1 + 3i \\ z^{2} = (1 + 3i)^{2} \\ z^{2} = (1 + 3i)(1 + 3i) \\ z^{2} = 1.1 + 1.3i + 3i.1 + 9 i^{2} \\ z^{2} = 1 + 3i + 3i + 9 i^{2} \\ z^{2} = 9 i^{2} + 6i + 1 \\ i^{2} = -1 \\ z^{2} = 9.(-1) + 6i + 1 \\ z^{2} = -9 + 6i + 1 \\ z^{2} = -8 + 6i z = 1 + 3i \\ z^{2} = (1 + 3i)^{2} \\ z^{2} = (1 + 3i)(1 + 3i) \\ z^{2} = 1.1 + 1.3i + 3i.1 + 9 i^{2} \\ z^{2} = 1 + 3i + 3i + 9 i^{2} \\ z^{2} = 9 i^{2} + 6i + 1 \\ i^{2} = -1 \\ z^{2} = 9.(-1) + 6i + 1 \\ z^{2} = -9 + 6i + 1 \\ z^{2} = -8 + 6i](https://tex.z-dn.net/?f=z+%3D+1+%2B+3i+%5C%5C++z%5E%7B2%7D+%3D++%281+%2B+3i%29%5E%7B2%7D++%5C%5C++z%5E%7B2%7D+%3D+%281+%2B+3i%29%281+%2B+3i%29+%5C%5C++z%5E%7B2%7D+%3D+1.1+%2B+1.3i+%2B+3i.1+%2B+9+i%5E%7B2%7D++%5C%5C++z%5E%7B2%7D+%3D+1+%2B+3i+%2B+3i+%2B+9+i%5E%7B2%7D++%5C%5C++z%5E%7B2%7D+%3D+9+i%5E%7B2%7D+%2B+6i+%2B+1+%5C%5C++i%5E%7B2%7D+%3D+-1+%5C%5C++z%5E%7B2%7D+%3D+9.%28-1%29+%2B+6i+%2B+1+%5C%5C++z%5E%7B2%7D+%3D+-9+%2B+6i+%2B+1+%5C%5C++z%5E%7B2%7D+%3D%C2%A0-8+%2B+6i)
Espero ter ajudado.
Espero ter ajudado.
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