Actions with complex numbers.

Complex numbers are numbers of the form $x+iy$, where $x,y\in \mathbb{R}$, and $i$ is a number such that $i^2=-1$. The set of complex numbers is denoted by $\mathbb{C}$.

Operations with complex numbers.

Addition of complex numbers:

$$(x_1+iy_1)+(x_2+iy_2)=(x_1+x_2)+i(y_1+y_2).$$

Multiplication of two complex numbers:

$$(x_1+iy_1)(x_2+iy_2)=x_1x_2-y_1y_2+(x_1y_2+x_2y_1)i.$$

Multiplication of a complex number by a real number:

$$\lambda(x+iy)=\lambda x+i\lambda y.$$

Division of complex numbers:

$$\frac{x_1+iy_1}{x_2+iy_2}=\frac{(x_1+iy_1)(x_2-iy_2)}{(x_2+iy_2)(x_2-iy_2)}=\frac{x_1x_2+y_1y_2+i(y_1x_2-x_1y_2)}{x_2^2+y_2^2}=$$ $$\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+\frac{y_1x_2-x_1y_2}{x_2^2+y_2^2}i.$$

The real numbers $x$ and $y$ of the complex number $z=x+iy$ are called the real and imaginary parts of the number $z$, respectively, and are denoted by $Re z=x$ and $Im z=y$.

Two complex numbers $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ are called equal if and only if $x_1=x_2$ and $y_1=y_2$.

The expression $z=x+iy$ is called the algebraic form of the complex number $z$.

The numbers $z_1=x+iy$ and $z_2=x-iy$ are called conjugate.

Examples:

Perform operations with complex numbers, presenting the result in algebraic form:

1.421. $(2+3i)(3-i).$

Solution:

$(2+3i)(3-i)=6-2i+9i-3i^2=6+7i+3=9+7i.$

Answer: $9+7i.$

1.424. $(2i-i^2)^2+(1-3i)^3.$

Solution:

$(2i-i^2)^2+(1-3i)^3=(2i+1)^2+1-3(3i)^2+3(3i)-(3i)^3=$ $=4i^2+4i+1-27i^2+9i-27i^3=-4+4i+1+27-9i+27i=24+22i.$

Answer: $24+22i.$

1.425. $\frac{2-i}{1+i}.$

Solution:

$$\frac{2-i}{1+i}=\frac{(2-i)(1-i)}{(1+i)(1-i)}=\frac{2-2i-i+i^2}{1-i^2}=\frac{2-3i-1}{1+1}=\frac{1-3i}{2}=\frac{1}{2}-\frac{3}{2}i.$$

Answer: $\frac{1}{2}-\frac{3}{2}i.$

1.428. $\frac{(1+i)(3+i)}{3-i}-\frac{(1-i)(3-i)}{3+i}.$

Solution.

$$\frac{(1+i)(3+i)}{3-i}-\frac{(1-i)(3-i)}{3+i}=\frac{(1+i)(3+i)(3+i)}{(3-i)(3+i)}-$$ $$-\frac{(1-i)(3-i)(3-i)}{(3+i)(3-i)}=\frac{9+15i+7i^2+i^3}{9-i^2}-\frac{9-15i+7i^2-i^3}{9-i^2}=$$ $$=\frac{9+15i-7-i-9+15i+7-i}{10}=\frac{28}{10}i=\frac{14}{5}i.$$

Answer: $\frac{14}{5}i.$

Find the real solutions of the following equation:

1. 430. $(1+i)x+(-2+5i)y=-4+17i.$

Solution.

$(1+i)x+(-2+5i)y=-4+17i\Rightarrow$

$x+xi-2y+5yi=-4+17i\Rightarrow$

$(x-2y)+(x+5y)i=-4+17i\Rightarrow$

$$\left\{\begin{array}{lcl}x-2y=-4\\x+5y=17\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}x=2y-4\\2y-4+5y=17\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}x=2\\y=3\end{array}\right. .$$

Answer: $x=2; y=3.$

Homework.

Perform operations with complex numbers, presenting the result in algebraic form:

1.422. $(1+2i)^2.$

Answer: $-3+4i.$

1.423. $(1-i)^3-(1+i)^3.$

Answer: $-4i.$

1.426. $\frac{1}{1+4i}+\frac{1}{4-i}.$

Answer: $\frac{5}{17}-\frac{3}{17}i.$

1.427. $\left(\frac{1-i}{1+i}\right)^3.$

Answer: $i.$

Find the real solutions of the following equation: