Question

Sa se det. Partea reala z=$\frac{1+i}{1-2i} + \frac{1+2i}{1-i}$

Answer 1

amplificam prima fractie cu (1-i) si a doua cu (1-2i)

(1-i)(1+i) + (1-2i)(1+2i)
--------------------------------
(1-2i)(1-i)

sus aplicam formula (a+b)(a-b)=a^2-b^2, adica
(1-i^1)+(1-4i^2)
---------------------
1-i-2i-2i^2

1-(-1)+(1-4(-1))
-------------------=
1-3i+2

1+1+(1+4)
--------------
3+3i

2+5
-------   = 
3(1+i)

7
----
3(1+i)

Answer 2

z = a + bi cu          Real(z) =a    si Im(z) =b
z = [ 1 +i  -i  -i² + 1 -2i +2i - 4i² ]  /  [ 1 - 2i -i  + 2 i² ]=
 = [ 1 + 1  +1  + 4 ] / [ 1 - 3i  -2 ] =  7 /  [ - 1 -3i ] = - 7  / [ 1 + 3i ] = 
= -7 · ( 1  -3i ) / [ ( 1 +3i ) · ( 1 -3i) =
= -7 ·( 1 -3i) / 10 
= - 7 / 10  + 21i/10 
Real( z)  = - 7 /10 
Im(z)  = 21/10