Asked by Lee
Will you explain the concept of "i" in algebra?
Problem: 4+2i/4-2i
Problem: 4+2i/4-2i
Answers
Answered by
Damon
i = sqrt (-1)
so i^2 = sqrt (-1)(sqrt(-1) = -1
(4+2i)/(4-2i)
to get rid of i in the denominator multiply top and bottom by the "comlex conjugate of (4-2i) which is (4+2i)
Rule: conjugate of (a-bi) is (a+bi)
(4+2i)(4+2i)
-------------
(4-2i)(4+2i)
16 + 16 i +4 i^2
=-------------------
(16 - 2 i^2)
but we all know that i^2 = -1 so
12 + 16 i
----------
18
or
6 + 8 i
--------
9
so i^2 = sqrt (-1)(sqrt(-1) = -1
(4+2i)/(4-2i)
to get rid of i in the denominator multiply top and bottom by the "comlex conjugate of (4-2i) which is (4+2i)
Rule: conjugate of (a-bi) is (a+bi)
(4+2i)(4+2i)
-------------
(4-2i)(4+2i)
16 + 16 i +4 i^2
=-------------------
(16 - 2 i^2)
but we all know that i^2 = -1 so
12 + 16 i
----------
18
or
6 + 8 i
--------
9
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