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Lösung 3.1:3

Aus Online Mathematik Brückenkurs 2

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<center> [[Image:3_1_3.gif]] </center>
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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 numerator:
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<math>\begin{align}\frac{3+i}{2+ai}&=\frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\
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&=\frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\
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&=\frac{6+a+(2-3a)i}{4+a^2}\\
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&=\frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}i.\end{align}</math>
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The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.
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NOTE: Think about how to solve the problem is a is not a real number.
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Version vom 10:29, 23. Sep. 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 numerator:


3+i2+ai=(3+i)(2−ai)(2+ai)(2−ai)=22−(ai)232−3ai+i2−ai2=4+a26+a+(2−3a)i=6+a4+a2+4+a22−3ai


The expression has real part equal to zero when 6+a=0, i.e. a=−6.

NOTE: Think about how to solve the problem is a is not a real number.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:3“