Processing Math: Done
Lösung 3.1:4b
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
Zeile 1: | Zeile 1: | ||
If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side, | If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side, | ||
- | {{ | + | {{Abgesetzte Formel||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}} |
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator, | It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt] | z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt] | ||
&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt] | &= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt] | ||
Zeile 15: | Zeile 15: | ||
Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly, | Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\text{LHS} | \text{LHS} | ||
&= (2-i)z\\[5pt] | &= (2-i)z\\[5pt] |
Version vom 13:06, 10. Mär. 2009
If we divide both sides by
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
![]() ![]() ![]() ![]() |
Also, we substitute
![]() ![]() ![]() ![]() ![]() ![]() |