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Maksim231197 [3]
3 years ago
7

Simplify (Are all division)2+4i3i3+2i4+i2i^11

Mathematics
1 answer:
Alla [95]3 years ago
6 0

We\ know:\ i=\sqrt{-1}\to i^2=-1

\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i

\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i

2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i


Used:\ (a+b)(a-b)=a^2-b^2

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