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Lösung 3.1:1d

(Unterschied zwischen Versionen)

Version vom 10:29, 11. Mär. 2009

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}