Question

Demonstrati ca 1+3i supra 1-3i + 1-3i supra 1+3i apartine lui R

Answer 1

$\dfrac{1+3i}{1-3i}+\dfrac{1-3i}{1+3i}=\dfrac{(1+3i)^2+(1-3i)^2}{(1-3i)(1+3i)}=\dfrac{1+6i-9+1-6i-9}{1+9}=\\ \dfrac{-16}{10}\in \mathbb{R}$

Answer 2

((1+3i)²+(1-3i)²)/(1+9)= ****(a+bi+a-bi)/10=2a/10=a/5 unde a=Re (1+3i)²,
deci a/5∈R, C.C.T.D.


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am tinut cont ca (zconjugat)^n=(z^n) conjugat;
deci daca (1+3i)²=a+bi, atunci (1-3i)²=a-bi
nu ne intereseaza cat sun a si b, stimca asi b∈R, cf.definitiei numerelor complexe sub forma algebrica