SOLUTION: Write 5/(6-2i) in a+bi form.

Question 72133This question is from textbook Advanced Algebra
: Write 5/(6-2i) in a+bi form. This question is from textbook Advanced Algebra

Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Multiply both top and bottom by the complex conjugate of 6-2i
FOIL the denominator (remember )
The i terms cancel in the denominator
Distribute the 5 among the parenthesis

Here's the answer in a+bi form where a=3/4 and b=1/4

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!

.
to get this to the form a + bi begin by multiplying the given term by:
.

.
Note that this is equivalent to multiplying the given term by 1 because
equals 1.
.
The numerator multiplication of the 5 times the (6 + 2i) results in 30 + 20i.
.
Then the denominator multiplication of results in:
.

.
The -12i and + 12i cancel each other out. Then recall that by definition is -1.
Substituting -1 for leads to:
.

.
This is the denominator ... +40. From above the numerator is 30 + 20i. So the answer is:
.

.
The answer to your problem is where and
.
Hope this helps you to understand complex numbers. Notice how you can eliminate complex
numbers in the denominator by multiplying the denominator by the same complex number with
a change in signs between the real and imaginary parts. This converts the denominator
to a real number.
.

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