Question

Ma puteți ajuta, va rog? ​

Answer 1

Bună! ✳

________

$^{1+3i)} \frac{1+3i}{1-3i} +^{1-3i)} \frac{1-3i}{1+3i} =$

$=\frac{(1+3i)^{2} }{(1-3i)(1+3i)} +\frac{(1-3i)^{2} }{(1+3i)(1-3i)} =$

$= \frac{1+6i+9i^{2}+1-6i+9i^{2} }{(1+3i)(1-3i)} =$

$=\frac{2-9-9}{1-9i^{2} } =$

$=\frac{-16}{1+9} =$

$=\frac{-16}{10} ^{(2} =-\frac{8}{5}$

$-\frac{8}{5}$ ∈ $R$ ✔

___________

① $i^{2} =-1$

② Când avem un număr complex la numitor, amplificăm fracția cu conjugatul acestuia.

z=a+bi → $\frac{}{z}$=a-bi

③ +6i-6i=0

Answer 2

Răspuns:

$-\frac{8}{5}$

Explicație pas cu pas:

$\frac{i+3i}{1-3i}+\frac{1-3i}{1+3i}=\frac{(i+3i)^{2}+(1-3i)^{2} }{(1-3i)(1+3i)}=\frac{i+6i+9i^{2}+1-6i+9i^{2} }{1-9i^{2} }=\frac{2+18i^{2} }{1-9i^{2} }$

prin definitie, $i^{2}=-1$

$<=>\frac{2+18*(-1)}{1-9*(-1)}=\frac{2-18}{1+9}=\frac{-16}{10}=-\frac{8}{5}$