Solution 3.3:3d
From Förberedande kurs i matematik 2
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| - | {{ | + | Before we can complete the square of the expression, we need to take out the factor |
| - | < | + | <math>i</math> in front of <math>z^2</math>, |
| - | {{ | + | |
| + | {{Displayed math||<math>i\Bigl(z^2+\frac{2+3i}{i}z-\frac{1}{i}\Bigr)\,\textrm{.}</math>}} | ||
| + | |||
| + | Then, simplify the complex fractions by multiplying top and bottom by <math>-i</math> (the denominator's complex conjugate), | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | i\Bigl(z^2+\frac{(2+3i)\cdot (-i)}{i\cdot (-i)}z-\frac{1\cdot (-i)}{i\cdot (-i)}\Bigr) | ||
| + | &= i\Bigl(z^2+\frac{-2i+3}{1}z-\frac{-i}{1}\Bigr)\\[5pt] | ||
| + | &= i\bigl(z^2+(3-2i)z+i\bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Now we are ready to complete the square of the second-degree expression inside the bracket, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | i\bigl(z^2+(3-2i)z+i\bigr) | ||
| + | &= i\Bigl(\Bigl(z+\frac{3-2i}{2}\Bigr)^2 - \Bigl(\frac{3-2i}{2}\Bigr)^2+i\Bigr)\\[5pt] | ||
| + | &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2 - \bigl(\tfrac{3}{2}-i\bigr)^2+i\bigr)\\[5pt] | ||
| + | &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2-\tfrac{9}{4}+3i-i^2+i\bigr)\\[5pt] | ||
| + | &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2-\frac{5}{4}+4i\bigr)\\[5pt] | ||
| + | &= i\bigl(z+\tfrac{3}{2}-i\bigr)^2-\tfrac{5}{4}i+4i^2\\[5pt] | ||
| + | &= i\bigl(z+\tfrac{3}{2}-i\bigr)^2-4-\tfrac{5}{4}i\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
Before we can complete the square of the expression, we need to take out the factor \displaystyle i in front of \displaystyle z^2,
| \displaystyle i\Bigl(z^2+\frac{2+3i}{i}z-\frac{1}{i}\Bigr)\,\textrm{.} |
Then, simplify the complex fractions by multiplying top and bottom by \displaystyle -i (the denominator's complex conjugate),
| \displaystyle \begin{align}
i\Bigl(z^2+\frac{(2+3i)\cdot (-i)}{i\cdot (-i)}z-\frac{1\cdot (-i)}{i\cdot (-i)}\Bigr) &= i\Bigl(z^2+\frac{-2i+3}{1}z-\frac{-i}{1}\Bigr)\\[5pt] &= i\bigl(z^2+(3-2i)z+i\bigr)\,\textrm{.} \end{align} |
Now we are ready to complete the square of the second-degree expression inside the bracket,
| \displaystyle \begin{align}
i\bigl(z^2+(3-2i)z+i\bigr) &= i\Bigl(\Bigl(z+\frac{3-2i}{2}\Bigr)^2 - \Bigl(\frac{3-2i}{2}\Bigr)^2+i\Bigr)\\[5pt] &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2 - \bigl(\tfrac{3}{2}-i\bigr)^2+i\bigr)\\[5pt] &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2-\tfrac{9}{4}+3i-i^2+i\bigr)\\[5pt] &= i\bigl(\bigl(z+\tfrac{3}{2}-i\bigr)^2-\frac{5}{4}+4i\bigr)\\[5pt] &= i\bigl(z+\tfrac{3}{2}-i\bigr)^2-\tfrac{5}{4}i+4i^2\\[5pt] &= i\bigl(z+\tfrac{3}{2}-i\bigr)^2-4-\tfrac{5}{4}i\,\textrm{.} \end{align} |
