Solution 3.1:1e

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A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it,
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<center> [[Image:3_1_1e.gif]] </center>
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{{Displayed math||<math>\begin{align}
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(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt]
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&= 4-4i+i^2\\[5pt]
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&= 4-4i-1\\[5pt]
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&= 3-4i\,\textrm{.}
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\end{align}</math>}}
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After that, we calculate the remaining product,
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{{Displayed math||<math>\begin{align}
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(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt]
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&= 3-4i+3i-4i^2\\[5pt]
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&= 3+(-4+3)i-4\cdot (-1)\\[5pt]
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&= 3-i+4\\[5pt]
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&= 7-i\,\textrm{.}
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\end{align}</math>}}

Current revision

A suitable first step can be to work out the square term, \displaystyle (2-i)^2, by expanding it,

\displaystyle \begin{align}

(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt] &= 4-4i+i^2\\[5pt] &= 4-4i-1\\[5pt] &= 3-4i\,\textrm{.} \end{align}

After that, we calculate the remaining product,

\displaystyle \begin{align}

(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt] &= 3-4i+3i-4i^2\\[5pt] &= 3+(-4+3)i-4\cdot (-1)\\[5pt] &= 3-i+4\\[5pt] &= 7-i\,\textrm{.} \end{align}

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