(3+4i)z(1-3i)=2+5i

2 min read Jun 16, 2024

Solving for Complex Number z: (3+4i)z(1-3i) = 2+5i

This problem involves solving for a complex number z in the equation (3+4i)z(1-3i) = 2+5i. We can achieve this by isolating z using complex arithmetic operations.

1. Expanding the Equation

First, expand the left side of the equation by multiplying the complex numbers:

(3+4i)z(1-3i) = (3+4i)(1-3i)z

Using the distributive property (or FOIL), we get:

(3+4i)(1-3i) = 3 - 9i + 4i - 12i²

Since i² = -1:

(3+4i)(1-3i) = 3 - 5i + 12 = 15 - 5i

Therefore, the equation becomes:

(15 - 5i)z = 2 + 5i

2. Isolating z

To isolate z, divide both sides of the equation by (15 - 5i):

z = (2 + 5i) / (15 - 5i)

3. Rationalizing the Denominator

To simplify the expression, we need to rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the denominator, which is (15 + 5i):

z = (2 + 5i) / (15 - 5i) * (15 + 5i) / (15 + 5i)

4. Simplifying the Expression

Expanding the multiplication and simplifying, we get:

z = (30 + 10i + 75i + 25i²) / (225 + 25i² )

Since i² = -1:

z = (30 + 85i - 25) / (225 - 25) = (5 + 85i) / 200

5. Expressing the Solution in Standard Form

Finally, we can express the solution in the standard form of a complex number:

z = 1/40 + 17/40i

Therefore, the complex number z that satisfies the equation (3+4i)z(1-3i) = 2+5i is z = 1/40 + 17/40i.