Answer:
1+√6+i(√3+√2)
3
Step-by-step explanation:
i will start from multiplying the numerator and denominator by the conjugate
(1+√3 i)(1+√2 i)
(1-√2 i)(1+√2 i)-----> 1²-(√2 i)²
open brackets
1+√3i+√2i+√6i²
1-2i²
i²= -1
1+i(√3+√2)+√6(-1)
1-2(-1)
1+√6+i(√3+√2)
3
quick clarification
[tex]\cfrac{1+\sqrt{3}i}{1-\sqrt{2}i}\cdot \cfrac{1+\sqrt{2}i}{1+\sqrt{2}i}\implies \cfrac{(1+\sqrt{3}i)(1+\sqrt{2}i)}{\underset{\textit{difference of squares}}{1^2-(\sqrt{2}i)^2}}\implies \cfrac{1+\sqrt{2}i+\sqrt{3}i+\sqrt{6}i^2}{1-[2i^2]} \\\\\\ \cfrac{1+\sqrt{2}i+\sqrt{3}i+\sqrt{6}(-1)}{1-[2(-1)]}\implies \cfrac{1-\sqrt{6}+i(\sqrt{2}+\sqrt{3})}{3}[/tex]
