Calculate the difference 1/(5+2i) - 1/(5-2i) .

hala718 | Certified Educator

Let z = 1/(5+2i)  - 1/(5-2i)

First le will determine common denominator.

(5+2i)(5-2i) = 25 + 4 = 29

==> z = [(5-2i) - (5+2i)]/29

= (5-2i-5-2i)/29

= -4i/29

==> z = -(4/29) i

william1941 | Student

The difference 1/ (5 + 2i) - 1 / ( 5 - 2i) can be solved as follows:

Multiply the numerator and denominator of first term with ( 5 - 2i) and that of the second term with ( 5 + 2i). Now the denominator is same for both and we have:

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

=> [5 - 5 - 2i - 2i ]/ ( 5^2 - 2^2*i^2)

=> ( - 4i) / (25 + 4)

=> - 4i / 29

Therefore the required result is -4i / 29

neela | Student

To calculate 1/(5+2i) - 1/(5+2i)

We rationalise the denominators by multiplying by their conjugates (here 5-2i is conjugate of 5+2i and 5+2i is the conjugate of 5-2i).

1/(5+2i) = (5-2i)/(5-2i)(5+2i) = (5-2i)/(5^2-4i^2) = ^5-2i)/(25+16), as i^2= -1.

Therefore 1/(5+2i) = (5-2i)/41.......(1)

Similarly we can show

-1/(5+2i) = - (5+2i)/41..................(2)

From (1) and (2) we get:

1/(5+2i) - 1/(5-2i) = (5-2i) /41 - (5+2i)/42 = {5-2i-5-2i}/41 = -4i/41

giorgiana1976 | Student

For the beginning, before calculating the difference of the 2 ratios, we have to transform the denominator of each ratio into a real number, instead of complex numbers.

We'll multiply the complex number from denominator by it's conjugate.

If the complex number is z=a+b*i, it's conjugate is z'=a-b*i.

So, if the complex number is 5+2i, it's conjugate is 5-2i.

We'll multiply the first ratio by the conjugate number (5-2i) and the second ratio by (5+2i).

(5-2i)/(5+2i)(5-2i) - (5+2i)/(5-2i)(5+2i) = (5-2i)/(25+4)-(5+2i)/(25+4)

(5-2i)/(25+4)-(5+2i)/(25+4) = (5-2i-5-2i)/29

We'll reduce like terms:

(5-2i-5-2i)/29= (-4i) / 29

So, the result of the difference between the 2 ratios is the complex number (-4i) / 29.

1/(5+2i) - 1/(5-2i) = (-4i) / 29