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Mathematics
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???
\[\frac{ 1+2i }{ \sqrt{2}+i }\]
Multiply both numerator and denominator by the conjugate of the denominator.I mean

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Other answers:

\[\sqrt{2}-i\]
\[\frac{1+2i}{\sqrt{2}+i}\cdot \frac{\sqrt{2}-i}{\sqrt{2}-i}\]
I got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2i ^{2} }{ 2-i ^{2}}\]
Great!
but also you know that \[i^2=-1\]
yes
so, it would be: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2(-1) }{ 2-(-1) }\]
yes, now just make arithmetic operations.
ok i got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}+2 }{ 3 }\] and this where i got stuck
is there a specific anwer you need to get?
yes i should get : \[\frac{ 2+\sqrt{2} }{ 3 }+\frac{ 2\sqrt{2}-1 }{ 3 }\]
I got: \[\frac{\sqrt{2}+2}{3}-\frac{1+2\sqrt{2}}{3}i\]
ok so how did you get that?

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