10.1 Given complex numbers $z_{1}=-2+5 i$ and $z_{2}=3-4 i$.

Find: a) $z_{1}+z_{2}$; b) $z_{2}-z_{1}$; c) $z_{1} z_{2}$; d) $\frac{z_{1}}{z_{2}}$.
10.1 a) Using formula (10.3), we get

$$
z_{1}+z_{2}=(-2+5 i)+(3-4 i)=(-2+3)+(5-4) i=1+i
$$

b) Similarly, according to formula (10.4), we find

$$
z_{2}-z_{1}=(3-4 i)-(-2+5 i)=(3+2)+(-4-5) i=5-9 i
$$

c) Using formula (10.5):

$$
\begin{aligned}
& z_{1} z_{2}=(-2+5 i)(3-4 i)= \\
& =(-2 \cdot 3-5(-4))+((-2)(-4)+5 \cdot 3) i=14+23 i
\end{aligned}
$$

The same result can be obtained by multiplying the numbers \(z_{1}\) and \(z_{2}\) according to the rule of multiplying binomials, taking into account the equality \(i^{2}=-1\).

d) To find the quotient \(\frac{z_{1}}{z_{2}}=\frac{-2+5 i}{3-4 i}\), we multiply the numerator and the denominator of the fraction by \(3+4 i\) - the number conjugate to the denominator. Then we get

$$
\begin{aligned}
& \frac{z_{1}}{z_{2}}=\frac{(-2+5 i)(3+4 i)}{(3-4 i)(3+4 i)}=\frac{-6+15 i-8 i-20}{3^{2}+4^{2}}= \\
& =\frac{-26+7 i}{25}=-\frac{26}{25}+\frac{7}{25} i
\end{aligned}
$$

Answer: a) \(1+i\); b) \(5-9 i\); c) \(14+23 i\); d) \(-1.04+0.28 i\).