From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
-	<center> [[Image:3_1_4f-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>z=x+iy</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_1_4f-2(2).gif]] </center>	+	and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
-	{{NAVCONTENT_STOP}}	+
		+	With this approach, the left-hand side of the equation becomes
		+
		+	{{Displayed math||<math>\begin{align}
		+	(1+i)(x-iy)+i(x+iy)
		+	&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
		+	&= x-iy+ix+y+ix-y\\[5pt]
		+	&= x+(2x-y)i
		+	\end{align}</math>}}
		+
		+	and the whole equation becomes
		+
		+	{{Displayed math||<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}
		+
		+	The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
		+
		+	{{Displayed math||<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}
		+
		+	This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
		+
		+	A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS}
		+	&= (1+i)\bar{z}+iz\\[5pt]
		+	&= (1+i)(3-i)+i(3+i)\\[5pt]
		+	&= 3-i+3i+1+3i-1\\[5pt]
		+	&= 3+5i\\[5pt]
		+	&= \text{RHS}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

Because the equation contains both \displaystyle z and \displaystyle \bar{z}, we cannot use \displaystyle z (or \displaystyle \bar{z}) alone as an unknown, so we are forced to set

\displaystyle z=x+iy

and use the real part \displaystyle x and the imaginary part \displaystyle y as unknowns.

With this approach, the left-hand side of the equation becomes

\displaystyle \begin{align} (1+i)(x-iy)+i(x+iy) &= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt] &= x-iy+ix+y+ix-y\\[5pt] &= x+(2x-y)i \end{align}

and the whole equation becomes

\displaystyle x+(2x-y)i=3+5i\,\textrm{.}

The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

\displaystyle \left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.

This gives \displaystyle x=3 and \displaystyle y=2x-5=2\cdot 3-5=1. Thus, the equation has the solution \displaystyle z=3+i.

A quick check shows that \displaystyle z=3+i satisfies the equation in the exercise,

\displaystyle \begin{align} \text{LHS} &= (1+i)\bar{z}+iz\\[5pt] &= (1+i)(3-i)+i(3+i)\\[5pt] &= 3-i+3i+1+3i-1\\[5pt] &= 3+5i\\[5pt] &= \text{RHS}\,\textrm{.} \end{align}