Finding the reciprical of imaginary numbers: -5-3i?

Answer 1

(-1) / (5 + 3 i)
(-1) (5 - 3i) / (5 + 3i)(5 - 3i)
(-1) (5 - 3i) / (25 + 9)
(-1/34) (5 - 3i)

Answer 2

Finding the reciprical of imaginary numbers: -5-3i
1 / (-5-3i) = (-5 +3i) /{(-5-3i)(-5 +3i)} =
(-5 +3i) / {25 +9} = (-5 +3i) /34 =
-5/34 + 3i /34

Answer 3

The reciprocal would have to be 1/(-5-3i), but this needs to be rationalized. First let's take a factor of -1 out and write it:

-1/(5+3i)

multiply numerator and denominator by the conjugate:

-1(5-3i) / [(5+3i)(5-3i)]

= (-5+3i)/(25+9)

= -5/34 + 3/34 i

that's it! :)

Answer 4

i^2 is comparable to -a million so... = ( 5 - 7i ) ( -4 - 3i ) = 5 ( -4 - 3i ) - 7i ( -4 - 3i ) = -20 - 15i + 28i + 21i^2 = -20 + 13i + 21(-a million) = -20 + 13i - 21 = -40-one + 13i x^2 - 22 = -112 x^2 = -112 + 22 x^2 = -ninety x = ?(-ninety) x = ?((3^2)(-10)) x = 3?(-10) x = 3?(10)i

Answer 5

-.147 + .088 i

multiply 1/(-5-3i) by (-5 + 3i)/(-5+3i)

= (-5 + 3i) / ( (-5 - 3i)*(-5+3i))
= (-5 + 3i) / (25 + 9)
= (-5 + 3i) / 34
= -5/34 + (3/34) *i
= -0.1470588 + .088235 *i

Answer 6

x * 1/x = 1 so

(a+bi)(-5-3i) = 1
-5a -5bi - 3ai - 3bi^2 = 1 + 0i
(-5a + 3b) - (5b + 3a)i = 1 + 0i

-5a + 3b = 1
5b + 3a = 0

3b = 1-5a
b = 1/3 - 5a/3

5(1/3 - 5a/3) + 3a = 0
5/3 + 25a/3 + 3a = 0
5 + 25a + 9a = 0
34a + 5 = 0
a = -5/34

b = 1/3 -5(5/34)/3
= 1/3 - 25/102
= (34-25)/102
= 9/102
= 3/34

1/(-5-3i) = -5/34 + 3/34 i