Homework Help

If the real part of complex z=1/(a(1+i)+1-2i) is 2/5, calculate a?

user profile pic

greenbel | Honors

Posted September 2, 2013 at 4:56 PM via web

dislike 1 like

If the real part of complex z=1/(a(1+i)+1-2i) is 2/5, calculate a?

1 Answer | Add Yours

Top Answer

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 2, 2013 at 5:49 PM (Answer #1)

dislike 1 like

You need to expand the complex number to denominator, such that:

`z = 1/(a + a*i + 1 - 2i)`

You need to group the members of denominator that contain i, such that:

`z = 1/(a + 1 + i*(a - 2))`

You need to multiplicate by the conjugate of denominator, such that:

`z = (a + 1 - i*(a - 2))/((a + 1 + i*(a - 2))(a + 1 - i*(a - 2)))`

`z = (a + 1 - i*(a - 2))/((a + 1)^2 - (a - 2)^2*i^2)`

Replacing -1 for `i^2` , yields:

`z = (a + 1 - i*(a - 2))/((a + 1)^2 + (a - 2)^2)`

`z = (a + 1 - i*(a - 2))/(a^2 + 2a + 1 + a^2 - 4a + 4)`

`z = (a + 1 - i*(a - 2))/(2a^2 - 2a + 5)`

You need to separate the complex number into its real part and imaginary part, such that:

`z = (a + 1)/(2a^2 - 2a + 5) - i*(a - 2)/(2a^2 - 2a + 5)`

Since the problem provides the information that the real part of complex number is `2/5` , you need to equate the real part of z and `2/5` , such that:

`(a + 1)/(2a^2 - 2a + 5) = 2/5 => 2(2a^2 - 2a + 5) = 5(a + 1)`

`4a^2 - 4a + 10 = 5a + 5 => 4a^2 - 9a + 5 = 0`

`a_(1,2) = (9 +- sqrt(81 - 80))/8`

`a_(1,2) = (9 +- sqrt1)/8 => a_(1,2) = (9 +- 1)/8`

`a_1 = 10/8 = 5/4`

`a_2 = 8/8 = 1`

Hence, evaluating a, under the given conditions, yields `a = 1` or `a = 5/4` .

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes