Solution 3.3:4c

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We complete the square on the left-hand side,
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<center> [[Image:3_3_4c.gif]] </center>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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(z+1)^2-1^2+3 &= 0\,,\\[5pt]
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(z+1)^2+2 &= 0\,\textrm{.}
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\end{align}</math>}}
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Taking the root now gives <math>z+1=\pm i\sqrt{2}</math>, i.e. <math>z=-1+i\sqrt{2}</math> and <math>z=-1-i\sqrt{2}</math>.
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We test the solutions in the equation to ascertain that we have calculated correctly.
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<math>\begin{align}
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z=-1+i\sqrt{2}:\quad z^2+2z+3
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&= \bigl(-1+i\sqrt{2}\,\bigr)^2 + 2\bigl(-1+i\sqrt{2}\bigr) + 3\\[5pt]
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&= (-1)^2 - 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 + 2i\sqrt{2} + 3\\[5pt]
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&= 1-2\cdot i\sqrt{2}-2-2+2i\sqrt{2}+3\\[5pt]
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&= 0,\\[10pt]
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z={}\rlap{-1-i\sqrt{2}:}\phantom{-1+i\sqrt{2}:}{}\quad z^2+2z+3
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&= \bigl(-1-i\sqrt{2}\,\bigr)^2 + 2\bigl(-1-i\sqrt{2}\,\bigr) + 3\\[5pt]
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&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
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&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
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&= 0\,\textrm{.}
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\end{align}</math>

Current revision

We complete the square on the left-hand side,

\displaystyle \begin{align}

(z+1)^2-1^2+3 &= 0\,,\\[5pt] (z+1)^2+2 &= 0\,\textrm{.} \end{align}

Taking the root now gives \displaystyle z+1=\pm i\sqrt{2}, i.e. \displaystyle z=-1+i\sqrt{2} and \displaystyle z=-1-i\sqrt{2}.

We test the solutions in the equation to ascertain that we have calculated correctly.

\displaystyle \begin{align} z=-1+i\sqrt{2}:\quad z^2+2z+3 &= \bigl(-1+i\sqrt{2}\,\bigr)^2 + 2\bigl(-1+i\sqrt{2}\bigr) + 3\\[5pt] &= (-1)^2 - 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 + 2i\sqrt{2} + 3\\[5pt] &= 1-2\cdot i\sqrt{2}-2-2+2i\sqrt{2}+3\\[5pt] &= 0,\\[10pt] z={}\rlap{-1-i\sqrt{2}:}\phantom{-1+i\sqrt{2}:}{}\quad z^2+2z+3 &= \bigl(-1-i\sqrt{2}\,\bigr)^2 + 2\bigl(-1-i\sqrt{2}\,\bigr) + 3\\[5pt] &= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt] &= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt] &= 0\,\textrm{.} \end{align}

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