Divide (4-5i)/(-5-6i) And explain? Ive seen the i disappears in some equations. I don't understand it and can't find any videos to help me?

(4 - 5i) / (-5 - 6i)

To rationalize the denominator, you have to multiply both halves of this fraction by the denominator's conjugate. In this case is (-5 + 6i)

After doing that and simplifying, your denominator will be a rational number:

(4 - 5i) / (-5 - 6i)
(4 - 5i)(-5 + 6i) / [(-5 - 6i)(-5 + 6i)]
(-20 - 24i + 25i - 30i²) / (25 - 30i + 30i - 36i²)
(-20 + i - 30i²) / (25 - 36i²)

Since i² = -1, make that substitution and simplify further:

(-20 + i + 30) / (25 + 36)
(10 + i) / 61

or:

10/61 + (1/61)i

Simplify (4-5i)/(-(6i)-5)
Factor -1 from -(6i)-5:
(4 -5i)/-(5 +6i)
Multiply numerator and denominator of (4-5i)/(-(5+6i)) by -1:
(-4 +5i))/(5+6i)
Multiply numerator and denominator by (1) as (5-6i):
(-(4-5 i) (5-6 i))/((5+6 i) (5-6 i))
use FOIL to multiply
(5+6 i) (5-6 i) = 25+36 = 61:
(4-5 i) (5-6 i) = 20 -24i -25i -30 = -10-49 i:
(--(49 i)-10)/61
Factor -1 from -(49 i)-10:
(--(10+49 i))/61
(-1)^2 = 1:

Answer: (10+49 i)/61

(4 - 5i) / (-5 -6i)
Multiply top and bottom by -5 + 6i.
(4 - 5i)(-5 + 6i) / (-5 + 6i)(-5 - 6i) =
(20 + 24i + 25i - 30i^2) / (25 - 36i^2) =
(50 + 49i) / 61

The trick to getting i out of the denominator is to multiply a + bi by a - bi.