Lösning 1.8.1b
Förberedande kurs i matematik
(Skillnad mellan versioner)
(Ny sida: <math>\displaystyle (1+2i)(2-\frac{i}{4}) = 1\cdot 2 + 1 \cdot \frac{i]{4}+2i\cdot 2+ 2i\cdot \frac{i}{4}</math>) |
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(3 mellanliggande versioner visas inte.) | |||
Rad 1: | Rad 1: | ||
- | <math>\ | + | <math>\begin{align} (3-2i)(4+i-(6-2i)) &= (3-2i)(-2+3i)=\\&=(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i =\\&= -6 +9i + 4i+6=\\&=13i\end{align} |
Nuvarande version
\displaystyle \begin{align} (3-2i)(4+i-(6-2i)) &= (3-2i)(-2+3i)=\\&=(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i =\\&= -6 +9i + 4i+6=\\&=13i\end{align}