Solution 3.3:3d

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

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