Question

Simplify 3 divided by 5-6i

Answer 1

For this case we must simplify the following expression:

$\frac {3} {5-6i}$

We multiply by:

$\frac {5 + 6i} {5 + 6i}\\\frac {3} {5-6i} * \frac {5 + 6i} {5 + 6i} =\\\frac {3 (5 + 6i)} {(5-6i) (5 + 6i)} =\\\frac {3 (5 + 6i)} {5 * 5 + 5 * 6i-6i * 5- (6i) ^ 2} =\\\frac {3 (5 + 6i)} {25-36i ^ 2} =\\\frac {3 (5 + 6i)} {25-36 (-1)} =\\\frac {3 (5 + 6i)} {25 + 36} =\\\frac {3 (5 + 6i)} {61} =\\\frac {15 + 18i} {61}$

Answer:

$\frac {15 + 18i} {61}$

Answer 2

Answer:

$\frac{15}{61}+\frac{18}{61}i$

Step-by-step explanation:

$\frac{3}{5-6i}$

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:

$\frac{3}{5-6i} \cdot \frac{5+6i}{5+6i}$

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:

$\frac{3}{5-6i} \cdot \frac{5+6i}{5+6i}=\frac{3(5+6i)}{25-36i^2}$

i^2=-1

$\frac{15+18i}{25-36(-1)}$

$\frac{15+18i}{25+36}$

$\frac{15+18i}{61}$

$\frac{15}{61}+\frac{18}{61}i$