1. Find the complex number \( z \) that satisfies the equations

$$
|z+2i|=|z-4i| \quad \text{and} \quad |z-4|=1.
$$
Solution. Let $z=a+i b$. By substitution in the equations, we get

$$
|a+i b+2 i|=|a+i b-4 i| \quad \text { and } \quad|a+i b-4|=1
$$

which are equivalent to

$$
\begin{array}{lll}
|a+i(b+2)|=|a+i(b-4)| & \text { and } & |a-4+i b|=1 \\
\sqrt{a^{2}+(2+b)^{2}}=\sqrt{a^{2}+(b-4)^{2}} & \text { and } & \sqrt{(a-4)^{2}+b^{2}}=1 \\
a^{2}+b^{2}+4 b+4=a^{2}+b^{2}-8 b+16 & \text { and } & a^{2}-8 a+16+b^{2}=1 \\
12 b=12 & \text { and } & a^{2}-8 a+16 b^{2}=1 .
\end{array}
$$

Therefore, $b=1$, and 4 is a solution to the equation $a^{2}-8 a+16=0$. Hence, $a=4$, so $z=4+i$.