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.

$$(2+i)\cdot(2+3i)= (2\cdot2-3\cdot1)+(2\cdot3+1\cdot2)i=1+8i$$

$$ \displaystyle \begin{array}{rl} (1+8i)\cdot(4-6i)=&(1\cdot4-8\cdot(-6))+(8\cdot4+1\cdot(-6))\cdot and =\\ =&4+48+(32-6)\cdot i=52+26i \end{array}$$

$$$ \displaystyle z=1-4i \Rightarrow \left\{ \begin{array}{l} \bar{z}=1-(-4)i=1+4i \\ -z=-(1-4i)=-1+4i \end{array} \right. $$$

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

The same as in the previous lines.

For the first one:

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

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

For the second one:

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

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