Solution 3.3:3c
From Förberedande kurs i matematik 2
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| - | {{ | + | If we take the minus sign out in front of the whole expression, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>-\bigl(z^2+2iz-4z-1\bigr)\,,</math>}} |
| + | |||
| + | and collect together the first-degree terms, | ||
| + | |||
| + | {{Displayed math||<math>-\bigl(z^2+(-4+2i)z-1\bigr)\,,</math>}} | ||
| + | |||
| + | we can then complete the square of the expression inside the outer bracket, | ||
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| + | {{Displayed math||<math>\begin{align} | ||
| + | -\bigl(z^2+(-4+2i)z-1\bigr) | ||
| + | &= -\Bigl(\Bigl(z+\frac{-4+2i}{2}\Bigr)^2-\Bigl(\frac{-4+2i}{2}\Bigr)^2-1\Bigr)\\[5pt] | ||
| + | &= -\bigl((z-2+i)^2-(-2+i)^2-1\bigr)\\[5pt] | ||
| + | &= -\bigl((z-2+i)^2-(-2)^2+4i-i^2-1\bigr)\\[5pt] | ||
| + | &= -\bigl((z-2+i)^2-4+4i+1-1\bigr)\\[5pt] | ||
| + | &= -\bigl((z-2+i)^2-4+4i\bigr)\\[5pt] | ||
| + | &= -(z-2+i)^2+4-4i\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
If we take the minus sign out in front of the whole expression,
| \displaystyle -\bigl(z^2+2iz-4z-1\bigr)\,, |
and collect together the first-degree terms,
| \displaystyle -\bigl(z^2+(-4+2i)z-1\bigr)\,, |
we can then complete the square of the expression inside the outer bracket,
| \displaystyle \begin{align}
-\bigl(z^2+(-4+2i)z-1\bigr) &= -\Bigl(\Bigl(z+\frac{-4+2i}{2}\Bigr)^2-\Bigl(\frac{-4+2i}{2}\Bigr)^2-1\Bigr)\\[5pt] &= -\bigl((z-2+i)^2-(-2+i)^2-1\bigr)\\[5pt] &= -\bigl((z-2+i)^2-(-2)^2+4i-i^2-1\bigr)\\[5pt] &= -\bigl((z-2+i)^2-4+4i+1-1\bigr)\\[5pt] &= -\bigl((z-2+i)^2-4+4i\bigr)\\[5pt] &= -(z-2+i)^2+4-4i\,\textrm{.} \end{align} |
