Solution 3.2:2b
From Förberedande kurs i matematik 2
| Line 4: | Line 4: | ||
<math>\begin{align} | <math>\begin{align} | ||
0 &\leq \mathrm{Re}z \leq 1,\\ | 0 &\leq \mathrm{Re}z \leq 1,\\ | ||
| - | 0 &\leq \mathrm{Im}z \leq 1,\end{align} | + | 0 &\leq \mathrm{Im}z \leq 1,\end{align}</math> |
| - | \mathrm{Re}z \leq \mathrm{Im}z | + | <math>\mathrm{Re}z \leq \mathrm{Im}z |
| - | </math> | + | </math>. |
The first two inequalities in this list define the unit square in the complex number plane. | The first two inequalities in this list define the unit square in the complex number plane. | ||
Revision as of 10:49, 3 October 2008
The inequality \displaystyle 0\leq \mathrm{Re}z \leq \mathrm{Im}z \leq 1 is actually several inequalities:
\displaystyle \begin{align} 0 &\leq \mathrm{Re}z \leq 1,\\ 0 &\leq \mathrm{Im}z \leq 1,\end{align}
\displaystyle \mathrm{Re}z \leq \mathrm{Im}z .
The first two inequalities in this list define the unit square in the complex number plane.
The last inequality says that the real part of \displaystyle z should be less than or equal to the imaginary part of \displaystyle z, I.e. \displaystyle z should lie to the left of the line \displaystyle y=x if \displaystyle x=\mathrm{Re} z and \displaystyle y = \mathrm{Im} z.
All together, the inequalities define the region which the unit square and the half-plane have in common: a triangle with corner points at \displaystyle 0, \displaystyle i and \displaystyle 1+i.
