Write 5-2i/3+2i in the form of a+bi where a and b are real numbers.?

Question

Write 5-2i/3+2i in the form of a+bi where a and b are real numbers.?

Answer 1

To do this, you combine the terms with an i and combine the terms with no i.

Since there is only one term with no i, a = 5

-(2/3)i + 2i = (-2/3)i + (6/3)i = (4/3)i

So b = 4/3

The answer is then

5 + (4/3)i

Answer 2

cant figure it out.  tryed

Answer 3

To divide two complex numbers, multiply the quotient by

the the form of ONE made from the complex conjugate of

the denominator, namely (3- 2i)/(3 - 2i)

That produces (5-2i)(3-2i)/(3+2i)(3-2i)

which gives [ 15 -  10i - 6i + 4i²]/ [9 - 4i²]

and that can be simplified by collecting like terms

in the numerator and replacing all i² with -1

to become [15 - 16i + 4(-1)]/[9 - 4(-1)]

= (11 -16i)/13  = (11/13) -  (16/13)i

Answer 4

I used the calculator to divide (5-2i) by (3+2i).

0.846153846 - 1.230769231 i

Answer 5

(5 - 2i)/(3 + 2i)

Multiply this by: (3 - 2i)/(3 - 2i)

[(5 - 2i)(3 - 2i)]/[(3 + 2i)(3 - 2i)]

The denominator: (3 + 2i)(3 - 2i) = 9 + 2i - 2i - 4i^2

Remembering that i^2 = -1:

9 + 4 = 13

The numerator: (5 - 2i)(3 - 2i) = 15 - 10i - 6i + 4i^2 = 11 - 16i

(5 - 2i)/(3 + 2i) = (11/13) - (16i/13)