Lösung 3.1:3
Aus Online Mathematik Brückenkurs 2
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| - | + | In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator, | |
| - | In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the | + | |
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| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{3+i}{2+ai} | ||
| + | &= \frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\[5pt] | ||
| + | &= \frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\[5pt] | ||
| + | &= \frac{6+a+(2-3a)i}{4+a^2}\\[5pt] | ||
| + | &= \frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}\,i\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>. | The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>. | ||
| - | NOTE: Think about how to solve the problem if a is not a real number. | ||
| - | + | Note: Think about how to solve the problem if <math>a</math> is not a real number. | |
Version vom 07:11, 30. Okt. 2008
In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator,
| \displaystyle \begin{align}
\frac{3+i}{2+ai} &= \frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\[5pt] &= \frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\[5pt] &= \frac{6+a+(2-3a)i}{4+a^2}\\[5pt] &= \frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}\,i\,\textrm{.} \end{align} |
The expression has real part equal to zero when \displaystyle 6+a=0, i.e. \displaystyle a=-6.
Note: Think about how to solve the problem if \displaystyle a is not a real number.
