Question

Z1 = 1-2i z2=4-i Найти: а) модули комплексных чисел z1 и z2 б) z1+z2 в) z1-z2 г) z1*z2 д) z1:z2

Answer 1

А) z1 = |1-2i| = $\sqrt{1^2+(-2)^2}$ = $\sqrt{5}$
    z2 = |4-i| = $\sqrt{4^2+(-1)^2}$ = $\sqrt{17}$
б)|z1+z2| = |(1-2i)+(4-i)| = |5-3i| = $\sqrt{5^2+(-3)^2}$ = $\sqrt{34}$
в)|z1-z2| = |(1-2i)-(4-i)| = |-3-i| = $\sqrt{(-3)^2+(-1)^2}$ = $\sqrt{10}$
г)|z1*z2| = |(1-2i)(4-i)| = |4-i-8i+2i^2| = |4 -9i -2| = |2-9i| = $\sqrt{2^2+(-9)^2}$ = $\sqrt{85}$ 
д)|z1:z2| = |$\frac{1-2i}{4-i}$| = |$\frac{1-2i}{4-i} * \frac{4+i}{4+i}$| = |$\frac{(1-2i)(4+i)}{16-i^2}| = |\frac{6-7i}{17}| = |\frac{6}{17}- \frac{7}{17}i$| = $\sqrt{(\frac{6}{17})^2+ (\frac{7}{17})^2} = \sqrt{ \frac{36}{289}+ \frac{49}{289}} = \frac{ \sqrt{85}}{17}$