Solution 3.1:3

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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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Revision as of 10:29, 23 September 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:


\displaystyle \begin{align}\frac{3+i}{2+ai}&=\frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\ &=\frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\ &=\frac{6+a+(2-3a)i}{4+a^2}\\ &=\frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}i.\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 is a is not a real number.

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