Solution 3.3:5a

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Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the square root.
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<center> [[Image:3_3_5a.gif]] </center>
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{{NAVCONTENT_STOP}}
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We complete the square on the left-hand side,
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{{Displayed math||<math>\begin{align}
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(z-(1+i))^2-(1+i)^2+2i-1 &= 0\,,\\[5pt]
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(z-(1+i))^2-(1+2i+i^2)+2i-1&=0\,,\\[5pt]
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(z-(1+i))^2-1-2i+1+2i-1 &= 0\,,\\[5pt]
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(z-(1+i))^2-1 &= 0\,\textrm{.}
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\end{align}</math>}}
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Now, we see that the equation has the solutions
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{{Displayed math||<math>z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}
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&2+i\,,\\
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&i\,\textrm{.}
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\end{align}\right.</math>}}
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We test the solutions,
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<math>\begin{align}
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z=2+i:\quad z^2-2(1+i)z+2i-1
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&= (2+i)^2 - 2(1+i)(2+i)+2i-1\\[5pt]
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&= 4+4i+i^2-2(2+i+2i+i^2)+2i-1\\[5pt]
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&= 4+4i-1-4-6i+2+2i-1\\[5pt]
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&= 0\,,\\[10pt]
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z={}\rlap{i:}\phantom{2+i:}{}\quad z^2-2(1+i)z+2i-1
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&= i^2-2(1+i)i+2i-1\\[5pt]
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&= -1-2(i+i^2)+2i-1\\[5pt]
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&= -1-2i+2+2i-1\\[5pt]
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&= 0\,\textrm{.}
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\end{align}</math>

Current revision

Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the square root.

We complete the square on the left-hand side,

\displaystyle \begin{align}

(z-(1+i))^2-(1+i)^2+2i-1 &= 0\,,\\[5pt] (z-(1+i))^2-(1+2i+i^2)+2i-1&=0\,,\\[5pt] (z-(1+i))^2-1-2i+1+2i-1 &= 0\,,\\[5pt] (z-(1+i))^2-1 &= 0\,\textrm{.} \end{align}

Now, we see that the equation has the solutions

\displaystyle z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}

&2+i\,,\\ &i\,\textrm{.} \end{align}\right.

We test the solutions,

\displaystyle \begin{align} z=2+i:\quad z^2-2(1+i)z+2i-1 &= (2+i)^2 - 2(1+i)(2+i)+2i-1\\[5pt] &= 4+4i+i^2-2(2+i+2i+i^2)+2i-1\\[5pt] &= 4+4i-1-4-6i+2+2i-1\\[5pt] &= 0\,,\\[10pt] z={}\rlap{i:}\phantom{2+i:}{}\quad z^2-2(1+i)z+2i-1 &= i^2-2(1+i)i+2i-1\\[5pt] &= -1-2(i+i^2)+2i-1\\[5pt] &= -1-2i+2+2i-1\\[5pt] &= 0\,\textrm{.} \end{align}

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