SOLUTION: Hi, Can you please help with imaginary numbers? . Simplify . {{{ (3+4i)/(6-2i) }}} . Put in complex number form {{{ a + bi }}} . Thanks ahead of time, Levi

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi, Can you please help with imaginary numbers? . Simplify . {{{ (3+4i)/(6-2i) }}} . Put in complex number form {{{ a + bi }}} . Thanks ahead of time, Levi      Log On


   

Question 159832: Hi, Can you please help with imaginary numbers?
.
Simplify
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+%283%2B4i%29%2F%286-2i%29+
.
Put in complex number form +a+%2B+bi+
.
Thanks ahead of time, Levi
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+%283%2B4i%29%2F%286-2i%29+ Start with the given expression

+%28%283%2B4i%29%2F%286-2i%29%29%28%286%2B2i%29%2F%286%2B2i%29%29+ Multiply both numerator and denominator by conjugate of the denominator 6-2i


+%28%283%2B4i%29%286%2B2i%29%29%2F%28%286-2i%29%286%2B2i%29%29 Combine the fractions


+%2818%2B30i%2B8i%5E2%29%2F%2836-4i%5E2%29 FOIL


+%2818%2B30i-8%29%2F%2836%2B4%29 Replace i%5E2 with -1 and multiply


+%2810%2B30i%29%2F%2840%29 Combine like terms.


10%2F40%2B%2830%2F40%29i Break up the fraction


1%2F4%2B%283%2F4%29i Reduce


So +%283%2B4i%29%2F%286-2i%29+ simplifies to 1%2F4%2B%283%2F4%29i. In other words, +%283%2B4i%29%2F%286-2i%29+=+1%2F4%2B%283%2F4%29i. Notice how the result is in standard form a%2Bbi where a=1%2F4 and b=3%2F4