Answer:
[tex]\frac{15}{61}+\frac{18}{61}i[/tex]
Step-by-step explanation:
[tex]\frac{3}{5-6i}[/tex]
To simplify or to write in the form a+bi, you will need multiply the top and bottom by the bottom's conjugate like so:
[tex]\frac{3}{5-6i} \cdot \frac{5+6i}{5+6i}[/tex]
Keep in mind when multiplying conjugates you only have to multiply first and last.
That is the product of (a+b) and (a-b) is (a+b)(a-b)=a^2-b^2.
(a+b) and (a-b) are conjugates
Let's multiply now:
[tex]\frac{3}{5-6i} \cdot \frac{5+6i}{5+6i}=\frac{3(5+6i)}{25-36i^2}[/tex]
i^2=-1
[tex]\frac{15+18i}{25-36(-1)}[/tex]
[tex]\frac{15+18i}{25+36}[/tex]
[tex]\frac{15+18i}{61}[/tex]
[tex]\frac{15}{61}+\frac{18}{61}i[/tex]
For this case we must simplify the following expression:
[tex]\frac {3} {5-6i}[/tex]
We multiply by:
[tex]\frac {5 + 6i} {5 + 6i}\\\frac {3} {5-6i} * \frac {5 + 6i} {5 + 6i} =\\\frac {3 (5 + 6i)} {(5-6i) (5 + 6i)} =\\\frac {3 (5 + 6i)} {5 * 5 + 5 * 6i-6i * 5- (6i) ^ 2} =\\\frac {3 (5 + 6i)} {25-36i ^ 2} =\\\frac {3 (5 + 6i)} {25-36 (-1)} =\\\frac {3 (5 + 6i)} {25 + 36} =\\\frac {3 (5 + 6i)} {61} =\\\frac {15 + 18i} {61}[/tex]
Answer:
[tex]\frac {15 + 18i} {61}[/tex]
