Ответ:

1) z1+z2=(2–3i)+(4+3i)=2–3i+4+3i=6;

2) z1–z2=(2–3i)–(4+3i)=2–3i–4–3i=–2–6i;

3) z1·z2=(2–3i)·(4+3i)=2·4–3i·4+2·(3i)–(3i)·(3i)=8–12i+6i+9=17–6i ;

4) z1:z2=2−3i4+3i=(2−3i)⋅(4−3i)(4+3i)(4−3i)=

=2⋅4−3i⋅4−2⋅3i−3i⋅3i)42−(3i)2==

=8−12i−6i+9)16+9=17−18i25

5) решить квадратное уравнение .

a)

x2+8x+20=0

D=82–4·20=64–80=–16

√D=√16·√–1=4i

x1=(–8–4i)/2=–4–2i; x2=(–8+4i)/2=–4+2i

б)

25x2+12x+4=0

D=122–4·25·4=–256

√D=√256·√–1=16i

x1=(–12–16i)/50=(–6–8i)/25; x2=(–12+16i)/50=(–6+8i)/25;]