Solve for z if 2z-z' = (3-5i)/(2-3i)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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Let the complex number z= a + ib

z' = a - ib

2z - z' = (3-5i)/(2-3i)

=> (3 - 5i)(2 + 3i)/(4 + 9)

=> (6 - 10i + 9i - 15i^2)/13

=> (6 + 15 - i)/13

=> 21/13 - i/13

2a + 2ib - a + ib = 21/13 - i/13

=> a + 3ib = 21/13 - i/1321/13 - i/13

=> a = 21/13 and b = -1/39

The complex number z = 21/13 - i/39

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll consider the complex number put in rectangular form:

z = x + i*y. It's conjugate is z' = x - i*y

2z - z' = 2x + 2i*y - x + i*y

2z - z' = x + 3i*y (1)

We'll manage the right side and we'll multiply the fraction by the conjugate of the denominator.

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

(3-5i)(2+3i)/(2^2 + 3^2) = (6 + 9i - 10i + 15)/13

(3-5i)/(2-3i) = (21-i)/13 (2)

We'll put (1)=(2):

x + 3i*y = (21-i)/13

13x + 39i*y = 21 - i

Comparing, we'll get:

13x = 21

x = 21/13

y = -1/39

The complex number z is: z = (21/13) - (1/39)*i

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