Lösung 3.1:4f
Aus Online Mathematik Brückenkurs 2
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Version vom 10:33, 11. Mär. 2009
Because the equation contains both \displaystyle z and \displaystyle \bar{z}, we cannot use \displaystyle z (or \displaystyle \bar{z}) alone as an unknown, so we are forced to set
| \displaystyle z=x+iy |
and use the real part \displaystyle x and the imaginary part \displaystyle y as unknowns.
With this approach, the left-hand side of the equation becomes
| \displaystyle \begin{align}
(1+i)(x-iy)+i(x+iy) &= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt] &= x-iy+ix+y+ix-y\\[5pt] &= x+(2x-y)i \end{align} |
and the whole equation becomes
| \displaystyle x+(2x-y)i=3+5i\,\textrm{.} |
The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
| \displaystyle \left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right. |
This gives \displaystyle x=3 and \displaystyle y=2x-5=2\cdot 3-5=1. Thus, the equation has the solution \displaystyle z=3+i.
A quick check shows that \displaystyle z=3+i satisfies the equation in the exercise,
| \displaystyle \begin{align}
\text{LHS} &= (1+i)\bar{z}+iz\\[5pt] &= (1+i)(3-i)+i(3+i)\\[5pt] &= 3-i+3i+1+3i-1\\[5pt] &= 3+5i\\[5pt] &= \text{RHS}\,\textrm{.} \end{align} |
