Operations with complex numbers in binomic form

1. $(3+2i)+(8-2i)=(3+8)+(2+(-2))i=11+0i=11$

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.

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

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

1. With the formula of the product of complex numbers in binomic form: $$\displaystyle \begin{array}{rl} (3-6i)\cdot(1+2i) &= (3\cdot 1+6\cdot2) +(3\cdot2-6\cdot1)i= (3+12)+(6-6)i \\ & =15+0i= 15 \end{array} $$
2. 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}$

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

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.

2. Now, multiplying by the conjugate of $2 + 3i$ (which is $2-3i$) we have: $$\dfrac{-11+29i}{2+3i}=\dfrac{-11+29i}{2+3i}\cdot \dfrac{2-3i}{2-3i}$$ We do both products: $$\dfrac{-11+29i}{2+3i}\cdot \dfrac{2-3i}{2-3i}=\dfrac{(-11\cdot2-29\cdot3)+(-11\cdot3+29\cdot2)i}{2^2+3^2}$$ Joining terms and adding: $$\dfrac{(-11\cdot2-29\cdot3)+(-11\cdot3+29\cdot2)i}{2^2+3^2}=\dfrac{65+91i}{4+9}$$ If we separate the fraction in two terms we obtain: $$\dfrac{-11+29i}{2+3i}=\dfrac{65}{13}+\dfrac{91}{13}i$$ Simplified, that will be: $$\dfrac{-11+29i}{2+3i}=\dfrac{65}{13}+\dfrac{91}{13}i=5+7i$$