Монгол Бодлогын Сан

Заавар: $$\dfrac{a+bi}{c+di}=\dfrac{(a+bi)(c-di)}{(c+di)(c-di)}=\dfrac{ac+bd}{c^2+d^2}+\dfrac{bc-ad}{c^2+d^2}i$$

$a>0$ тоо үед сөрөг тооны арифметик язгуурыг $\sqrt{-a}=\sqrt{a}i$ гэж тодорхойлно. $$\sqrt{-9}\cdot\sqrt{-16}=\sqrt{9}i\cdot\sqrt{16}i=3i\cdot 4i=-12\neq\sqrt{(-9)\cdot(-16)}=12$$ тул эерэг бодит тоонуудын хувьд биелдэг $$\sqrt{a\vphantom{b}}\cdot\sqrt{b}=\sqrt{ab}$$ чанар сөрөг тоонууд дээр биелэхгүйг анхаар!

Бодолт:

1. $\dfrac{3+2i}{2+3i}=\dfrac{3\cdot 2+2\cdot 3}{2^2+3^2}+\dfrac{2\cdot 2-3\cdot 3}{2^2+3^2}i=\dfrac{12}{13}-\dfrac{5}{13}i$
2. 
   \begin{align*} \dfrac{2-i}{3+i}-\dfrac{5+10i}{1-3i}&=\left(\dfrac{2\cdot 3+(-1)\cdot 1}{3^2+1^2}+\dfrac{(-1)\cdot 3-2\cdot 1}{3^2+1^2}i\right)- \left(\dfrac{5\cdot 1+10\cdot (-3)}{1^2+(-3)^2}+\dfrac{10\cdot 1-5\cdot (-3)}{1^2+(-3)^2}i\right)\\ &=\left(\dfrac{5}{10}-\dfrac{5}{10}i\right)-\left(-\dfrac{25}{10}+\dfrac{25}{10}i\right)=\dfrac{5-(-25)}{10}-\dfrac{5+25}{10}i=3-3i \end{align*}
3. $\sqrt{-9}+\sqrt{-16}=3i+4i=7i$
4. $(5+\sqrt{-3})(4-\sqrt{-3})=(5+\sqrt3i)(4-\sqrt{3}i)=20-5\sqrt{3}i+4\sqrt{3}i-(\sqrt{3}i)^2=23-\sqrt3i$