Solving Complex Number Equations?

Question

Can somebody explain how to solve the equation pls.

1/R= 1/(2-5i) - 1/(1+4i)

Answer 1

First things first; your final answer should ultimately be in the form R = a + bi. As for how you'd solve it, you would first try and solve for R using the normal algebraic methods.

1/R = 1/(2 - 5i) - 1/(1 + 4i)

To get rid of all fractions, multiply everything by R(2 - 5i)(1 + 4i).

(2 - 5i)(1 + 4i) = R(1 + 4i) - R(2 - 5i)

Expand the left hand side, keeping in mind that i^2 is equal to -1.

2 + 2(4i) - 1(5i) - 20i^2 = R(1 + 4i) - R(2 - 5i)

Simplify

2 + 8i - 5i - 20(-1) = R(1 + 4i) - R(2 - 5i)
2 + 3i + 20 = R(1 + 4i) - R(2 - 5i)
3i + 22 = R(1 + 4i) - R(2 - 5i)

Now, on the right hand side, you're going to factor out an R.

3i + 22 = R[1 + 4i - (2 - 5i)]

Simplify, to get

3i + 22 = R[1 + 4i - 2 + 5i]
3i + 22 = R[-1 + 9i]

And then divide both sides by -1 + 9i, to get

(3i + 22)/(-1 + 9i) = R

At this point, in order to put your answer in the form a + bi, you have to multiply the numerator and denominator by the conjugate of the denominator (-1 - 9i). The reason you want to do this is because you'll put the denominator in the form a^2 - b^2, which would effectively remove the i from the denominator and help you put the question in the form a + bi.

R = [(3i + 22) (-1 - 9i)] / [ (-1)^2 - (9i)^2]
R = [-3i - 27i^2 -22 - 198i] / [1 - 81i^2]
R = [-201i + 27 - 22] / [1 + 81]
R = [-201i + 5] / [82]
R = (-201/82)i + 5/82

Or, to put it in its proper form

R = 5/82 - (201/82)i

Answer 2

1/R = 1/(2-5i) - 1/(1+4i) = (2+5i)/29 - (1-4i)/17 = (5+201i)/493

R = 493/(5+201i) = 493(5-201i)/40,426 = (5-201i)/82 = 5/82 - (201/82)i

Answer 3

1/R = (5/493) + (201\ 493)i (common denominator)
find the inverse:
R = (5/82) - (201/82)i