Question

Помогите пожалуйста решить:
1) (3+i)+(2+3i)
2) (2+3i)-(3+i)
3) (2-3i)+(2-3i)
4) (2+3i)-(5-8i)
5) (2+3i)*(4+2i)
6) (2-3i)/(5+2i)​

Answer 1

Действия с комплексными числами

1) (3 + i) + (2 + 3i)  = 3 + i + 2 + 3i = 5 + 4i

2) (2 + 3i) - (3 + i)  = 2 + 3i - 3 - i = -1 + 2i

3) (2 - 3i) + (2 - 3i)  = 2 · (2 - 3i) = 4 - 6i

4) (2 + 3i) - (5 - 8i)  = 2 + 3i - 5 + 8i = -3 + 11i

5) (2 + 3i) * (4 + 2i)  = 8 + 4i + 12i + 6i² = 8 + 16i - 6 = 2 + 16i

$6)~~~\dfrac{(2-3i)}{(5+2i)}=\dfrac{(2-3i)(5-2i)}{(5+2i)(5-2i)}=\\\\~~~=\dfrac{10-4i-15i+6i^2}{25-4i^2}=\dfrac{10-19i-6}{25+4}=\\\\~~~=\dfrac{4-19i}{29}\boldsymbol{=\dfrac 4{29}-\dfrac {19}{29}i}$

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$i=\sqrt {-1}\\\\i^2=\Big(\sqrt{-1}\Big)^2=-1\\\\(a-b)(a+b)=a^2-b^2$