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 denominator,
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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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{{Displayed math||<math>\begin{align}
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\frac{3+i}{2+ai}
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&= \frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\[5pt]
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&= \frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\[5pt]
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&= \frac{6+a+(2-3a)i}{4+a^2}\\[5pt]
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&= \frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}\,i\,\textrm{.}
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\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>.
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NOTE: Think about how to solve the problem is a is not a real number.
 
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Note: Think about how to solve the problem if <math>a</math> is not a real number.

Current revision

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.

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