Найдите модуль комплексного числа z= (1-i)(3-i) / (1+i)(4-i) a) 15/17 b) 12/17 c) √170/17 d) 13/17 e) 3√13/17

[latex]z= \frac{(1-i)(3-i)}{(1+i)(4-i)} = \frac{3-i-3i-1}{4-i+4i+1} = \frac{2-4i}{5+3i} = \frac{10-12}{25+9} +( \frac{-20-6}{25+9} )i=- \frac{1}{17} - \frac{13}{17}i\\ \\ |z|= \sqrt{(- \frac{1}{17})^2+(- \frac{13}{17})^2} = \frac{ \sqrt{170} }{17} [/latex]

[latex]\displaystyle z= \frac{(1-i)(3-i)}{(1+i)(4-i)} = \frac{3-i-3i-1}{4-i+4i+1}= \frac{2-4i}{5+3i}= \\ \frac{(2-4i)(5-3i)}{(5+3i)(5-3i)}= \frac{10-6i-20i-12}{25+9}= \frac{-2-26i}{34}= -\frac{1}{17}- \frac{13}{17}i; \\ \\ |z|= \sqrt{\left(-\frac{1}{17}\right)^2+\left(-\frac{13}{17}\right)^2} = \frac{1}{17} \sqrt{1+13^2}=\frac{1}{17} \sqrt{170} [/latex] Ответ с)