Question

Pentru liceu te rog mersi......

Answer 1

$\displaystyle \mathtt{a) \frac{2+3i}{1-i}= \frac{(2+3i)(1+i)}{(1-i)(1+i)}= \frac{(2 \cdot 1-3 \cdot 1)+(2 \cdot 1+3 \cdot 1)i}{1^2+(-1)^2} = }\\ \\ \mathtt{= \frac{(2-3)+(2+3)i}{2}= \frac{-1+5i}{2}=- \frac{1}{2}+ \frac{5}{2}i }$

$\displaystyle \mathtt{b) \frac{2i}{2-i}= \frac{2i(2+i)}{(2-i)(2+i)}= \frac{4i+2i^2}{2^2+(-1)^2} = \frac{4i-2}{5} =- \frac{2}{5}+ \frac{4}{5}i }$

$\displaystyle \mathtt{c) \frac{1+i \sqrt{3} }{1-i \sqrt{3} }=\frac{\left(1+i\sqrt{3}\right)\left(1+i \sqrt{3}\right) }{\left(1-i \sqrt{3}\right)\left(1+i \sqrt{3}\right)}=} \\ \\ \mathtt{ =\frac{\left(1 \cdot 1- \sqrt{3}\cdot \sqrt{3}\right) +\left(1 \cdot \sqrt{3}+ \sqrt{3} \cdot 1{\right)i} }{1^2+\left(-\sqrt{3}\right)^2}=\frac{(1-3)+\left( \sqrt{3}+ \sqrt{3}\right)i }{4}=}\\ \\\mathtt{ = \frac{-2+2 \sqrt{3}i }{4}=\frac{2\left(-1+\sqrt{3}i\right) }{4}=\frac{-1+\sqrt{3}i }{2}=-\frac{1}{2}+ \frac{\sqrt{3} }{2}i}$

$\displaystyle \mathtt{d) \frac{a \sqrt{b}+b \sqrt{a}i }{b \sqrt{a}-a \sqrt{b}i}= \frac{\left(a \sqrt{b}+b \sqrt{a}i\right)\left(b \sqrt{a} +a \sqrt{b}i\right) }{\left(b \sqrt{a}-a \sqrt{b}i\right)\left(b \sqrt{a}+a \sqrt{b}i\right) } =}\\ \\ \mathtt{= \frac{\left(a \sqrt{b} \cdot b \sqrt{a} -b\sqrt{a}\cdot a \sqrt{b}\right)+\left(a \sqrt{b}\cdot a\sqrt{b}+b\sqrt{a}\cdot b\sqrt{a} \right)i}{\left(b \sqrt{a}\right)^2+\left(-a \sqrt{b}\right)^2 }= }\\ \\ \mathtt{= \frac{\left(a^2b+ab^2\right)i}{ab^2+a^2b}=i}$