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ssskrzat · Post autor: **ssskrzat** » 12 sty 2012, 20:16

\(a) |\frac{-4}{1+i \sqrt{3} }|
b) |-8 + 15i| \cdot Re \frac{3+7i}{-3-i}\)

irena · Post autor: **irena** » 12 sty 2012, 20:38

a)
\(z=\frac{-4}{1+i\sqrt{3}}\ \cdot\ \frac{1-i\sqrt{3}}{1-i\sqrt{3}}=\frac{-4(1-i\sqrt{3})}{1+3}=-1+i\sqrt{3}\\|z|=\sqrt{1+3}=\sqrt{4}=2\)

irena · Post autor: **irena** » 12 sty 2012, 20:42

b)
\(|-8+15i|=\sqrt{64+225}=\sqrt{289}=17\)

\(z=\frac{3+7i}{-3-i}\ \cdot\ \frac{-3+i}{-3+i}=\frac{-9+3i-21i-7}{9-i^2}=\frac{-16-18i}{10}\\Re(z)=-1,6\)

\(17\cdot(-1,6)=-27,2\)