From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	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,
-	<center> [[Image:3_1_3.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>.
		+
		+
		+	Note: Think about how to solve the problem if <math>a</math> is not a real number.

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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.