Solution 3.1:1c

From Förberedande kurs i matematik 2

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Complex numbers satisfy the same rules of arithmetic as ordinary numbers, with the addition that <math>i^2=-1</math>. The distributivity rule gives that
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<center> [[Image:3_1_1c.gif]] </center>
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{{Displayed math||<math>\begin{align}
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i(2+3i)
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&= i\cdot 2 + i\cdot 3i\\[5pt]
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&= 2i+3i^2\\[5pt]
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&= 2i+3\cdot (-1)\\[5pt]
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&= 2i-3\\[5pt]
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&= -3+2i\,\textrm{.}
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\end{align}</math>}}

Current revision

Complex numbers satisfy the same rules of arithmetic as ordinary numbers, with the addition that \displaystyle i^2=-1. The distributivity rule gives that

\displaystyle \begin{align}

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