(1+i)(1-i)(3+2i)(3-2i)([latex] \sqrt{3} [/latex]-i)([latex] \sqrt{3} [/latex]+I)(5+I)(5-2i)(4+2i)(4-i)

Question

(1+i)(1-i)(3+2i)(3-2i)([latex] \sqrt{3} [/latex]-i)([latex] \sqrt{3} [/latex]+I)(5+I)(5-2i)(4+2i)(4-i)

Математика

(1+i)(1-i) (3+2i)(3-2i) ([latex] \sqrt{3} [/latex]-i)([latex] \sqrt{3} [/latex]+I) (5+I)(5-2i) (4+2i)(4-i)

Гость

Ответ(ы) на вопрос:

Accepted Answer

[latex](1+i)(1-i)=1-i^2=1+1=2 \\\ (3+2i)(3-2i)=9-4i^2=9+4=13 \\\ ( \sqrt{3} -i)( \sqrt{3} +i)=3-i^2=3+1=4 \\\ (5+i)(5-2i)-25-10i+5i-2i^2=25+2-5i=27-5i \\\ (4+2i)(4-i)=16-4i+8i-2i^2=16+2+4i=18+4i[/latex]