Problems from Operations with complex numbers in binomic form

$(3+2i)+(8-2i)=(3+8)+(2+(-2))i=11+0i=11$

We have added the real parts $(3+8)$ and we have added the imaginary parts $(2+(-2))$. This last component is zero and therefore the solution is a real number.

$(3+2i)-(8-2i)=(3-8)+(2-(-2))i=-5+(2+2)i=-5+4i$

We substract real parts $(3-8)$ and the imaginary parts $(2-(-2))$. The real part result is $-5$ and the imaginary part result is $4$.

We will first do the product of the first two complex numbers and then multiply the result by the third complex number.

$(2+i)\cdot (2+3i)=(2\cdot 2-3\cdot 1)+(2\cdot 3+1\cdot 2)i=1+8i$

${\displaystyle \begin{array}{rl}(1+8i)\cdot (4-6i)=& (1\cdot 4-8\cdot (-6))+(8\cdot 4+1\cdot (-6))\cdot and=\\ =& 4+48+(32-6)\cdot i=52+26i\end{array}}$

$${\displaystyle z=1-4i\Rightarrow \{\begin{array}{l}\overline{z}=1-(-4)i=1+4i\\ -z=-(1-4i)=-1+4i\end{array}}$$

The first one corresponds to the conjugate and the second one to the opposite complex number. $${\displaystyle z=-9-5i\Rightarrow \{\begin{array}{l}\overline{z}=-9-(-5)i=-9+5i\\ -z=-(-9-5i)=9+5i\end{array}}$$

The same as in the previous lines.

For the first one:

$\overline{z}=1-(-4)i=1+4i$

$-z=-(1-4i)=-1+4i$

For the second one:

$\overline{z}=-9-(-5)i=-9+5i$

$-z=-(-9-5i)=9+5i$