Solution 3.1:1d

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We expand the expression by multiplying each term in the first bracket with every term in the second bracket,
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<center> [[Image:3_1_1d.gif]] </center>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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(3-2i)(7+5i)
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&= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt]
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&= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.}
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\end{align}</math>}}
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Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together,
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{{Displayed math||<math>\begin{align}
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(3-2i)(7+5i)
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&=21+15i-14i-10i^2\\[5pt]
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&=21+15i-14i-10\cdot(-1)\\[5pt]
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&=(21+10)+(15i-14i)\\[5pt]
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&=31+(15-14)i\\[5pt]
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&=31+i\,\textrm{.}
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\end{align}</math>}}

Current revision

We expand the expression by multiplying each term in the first bracket with every term in the second bracket,

\displaystyle \begin{align}

(3-2i)(7+5i) &= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt] &= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.} \end{align}

Then, we use that \displaystyle i^2=-1 and write the real and imaginary parts together,

\displaystyle \begin{align}

(3-2i)(7+5i) &=21+15i-14i-10i^2\\[5pt] &=21+15i-14i-10\cdot(-1)\\[5pt] &=(21+10)+(15i-14i)\\[5pt] &=31+(15-14)i\\[5pt] &=31+i\,\textrm{.} \end{align}

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