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Algebra II please help!!!?
1.simplify
(3+5i)/(-2+2i)
2.simplify
(6+2i)+-(11-9i)
3.simpify
(-4+i)(-4-i)
4.simplify
(3+2i)(1+4i)
5.simplify
(3+2i)/(1+4i)
anything helps thank you so much!!!
3 Answers
- ?Lv 74 years ago
1.
(3 + 5i)/(-2 + 2i)
= [(3 + 5i)(2i + 2)]/[(2i - 2)(2i + 2)]
= (16i - 4)/-8
= 1/2 - 2i
2.
(6 + 2i) + -(11 - 9i)
= -5 + 11i
3.
(-4 + i)(-4 - i)
= 17
4.
(3 + 2i)(1 + 4i)
= -5 + 14i
5.
(3 + 2i)/(1 + 4i)
= -5 + 14i
- icemanLv 74 years ago
1) (3 + 5i)/(-2 + 2i) * (-2 - 2i)/(-2 - 2i)
= (-6 - 6i - 10i + 10)/8
= (4 - 16i)/8
= 1/2 - 2i
- ?Lv 74 years ago
1) When dividing by complex number, like /(-2+2i) multiply by (1) as (-2-2i)/(-2-2i)
2) distribute the negative sign -(11-9i) = -11 +9i, then combine like items
3) use FOIL like normal
4) use FOIL like normal
5) When dividing by complex number, like /(1+4i) multiply by (1) as (1-4i)/(1-4i)
