Lösning 1.8.1b
Förberedande kurs i matematik
(Skillnad mellan versioner)
| Rad 1: | Rad 1: | ||
| - | <math>\displaystyle (1+2i)\left( 2-\frac{i}{4} \right) = 1\cdot 2 - 1 \cdot \frac{i}{4}+2i\cdot 2- 2i\cdot \frac{i}{4}</math> = 2-\frac{i}{4}+4i | + | <math>\displaystyle (1+2i)\left( 2-\frac{i}{4} \right) = 1\cdot 2 - 1 \cdot \frac{i}{4}+2i\cdot 2- 2i\cdot \frac{i}{4}=</math> |
| + | |||
| + | <math>\displaystyle = 2-\frac{i}{4}+4i+\frac{1}{2} = \frac{5}{2} + \frac{15 i} {4}</math> | ||
Versionen från 12 juni 2012 kl. 13.58
\displaystyle \displaystyle (1+2i)\left( 2-\frac{i}{4} \right) = 1\cdot 2 - 1 \cdot \frac{i}{4}+2i\cdot 2- 2i\cdot \frac{i}{4}=
\displaystyle \displaystyle = 2-\frac{i}{4}+4i+\frac{1}{2} = \frac{5}{2} + \frac{15 i} {4}
