Asked by Yin
Find result of imaginary unit i.
1. (3+4i^259)(2-3i^-22)=?
2. (-i^-38)/(5-4i^-27)=?
1. (3+4i^259)(2-3i^-22)=?
2. (-i^-38)/(5-4i^-27)=?
Answers
Answered by
Steve
well, we all know that i = √-1.
I assume you mean you want the expression rewritten as a standard complex number, a+bi.
Since
...
i^-3 = i
i^-2 = -1
i^-1 = -i
i^0 = 1
i^2 = -1
i^3 = -i
i^4 = 1
...
#1
i^259 = i*(4*64+3) = i^3 = -i
i^-22 = i^(-4*6+2) = -1
(3+4i^259)(2-3i^-22)
= (3-4i)(2+3)
= 15-20i
#2
(-i^-38)/(5-4i^-27)
= (-(-1))/(5-4(i))
= 1/(5-4i)
= (5+4i)/(5^2+4^2)
= 5/41 + 4/41 i
I assume you mean you want the expression rewritten as a standard complex number, a+bi.
Since
...
i^-3 = i
i^-2 = -1
i^-1 = -i
i^0 = 1
i^2 = -1
i^3 = -i
i^4 = 1
...
#1
i^259 = i*(4*64+3) = i^3 = -i
i^-22 = i^(-4*6+2) = -1
(3+4i^259)(2-3i^-22)
= (3-4i)(2+3)
= 15-20i
#2
(-i^-38)/(5-4i^-27)
= (-(-1))/(5-4(i))
= 1/(5-4i)
= (5+4i)/(5^2+4^2)
= 5/41 + 4/41 i
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