Question

Aflați partea imaginara a numărului

Answer 1

$z\ =\ \frac{1}{4+3i}+\frac{\left(2-i\right)^2}{1+i}-\frac{i}{4i-3}+\frac{6}{2-i}$

$Im\left(z\right)=Im\left(\frac{1}{4+3i}\right)+Im\left(\frac{\left(2-i\right)^2}{1+i}\right)-Im\left(\frac{i}{4i-3}\right)+Im\left(\frac{6}{2-i}\right)$

$Im\left(z\right)=Im\left(\frac{4-3i}{4^2-\left(3i\right)^2}\right)+Im\left(\frac{\left(2-i\right)^2\left(1-i\right)}{1^2-i^2}\right)-Im\left(\frac{i\left(4i+3\right)}{\left(4i\right)^2-3^2}\right)+Im\left(\frac{6\left(2+i\right)}{2^2-i^2}\right)$

$Im\left(z\right)=Im\left(\frac{4-3i}{16+9}\right)+Im\left(\frac{\left(4-4i-1\right)\left(1-i\right)}{1+1}\right)-Im\left(\frac{-4+3i}{-16-9}\right)+Im\left(\frac{12+6i}{4+1}\right)$

$Im\left(z\right)=-\frac{3}{25}+Im\left(\frac{\left(3-4i\right)\left(1-i\right)}{2}\right)-\left(-\frac{3}{25}\right)+\frac{6}{5}$

$Im\left(z\right)\ =\ -\frac{3}{25\ }+\left(-\frac{7}{2}\right)+\frac{3}{25}+\frac{6}{5}$

$\Rightarrow Im(z) = \dfrac{-35+12}{10} = -\dfrac{23}{10}$

Answer 2

$z=\frac{1}{4+3i}+\frac{(2-i)^2}{1+i}-\frac{i}{4i-3}+\frac{6}{2-i}$

$z=\frac{4-3i}{(4-3i)(4+3i)}+\frac{(4-4i+i^2)\cdot (1-i)}{(1-i)(1+i)}-\frac{i(-4i-3)}{(-3+4i)(-3-4i)}+\frac{6(2+i)}{(2+i)(2-i)}$

$z=\frac{4-3i}{4^2-3^2i^2}+\frac{(4-4i-1)(1-i)}{1^2-i^2}-\frac{-4i^2-3i}{(-3)^2-4^2i^2}+\frac{12+6i}{2^2-i^2}$

$z=\frac{4-3i}{16+9}+\frac{(3-4i)(1-i)}{1+1}-\frac{-4\cdot (-1)-3i}{9+16}+\frac{12+6i}{4+1}$

$z=\frac{4-3i}{25}+\frac{3-3i-4i+4i^2}{2}-\frac{4-3i}{25}+\frac{12+6i}{5}$

$z=\frac{3-4-7i}{2}+\frac{12+6i}{5}$

$z=\frac{5(-1-7i)}{10}+\frac{2(12+6i)}{10}$

$z=\frac{-5-35i+24+12i}{10}$

$z=\frac{19-23i}{10}$

$z=\frac{19}{10}+\frac{-23}{10}i$

$\Rightarrow Im(z)=\frac{-23}{10}$