(2-3i)z-(9-2i)=(1+i)z

Jasonbradley

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Jun 16, 2024

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Solving Complex Equations: (2-3i)z - (9-2i) = (1+i)z

This article will guide you through solving the complex equation (2-3i)z - (9-2i) = (1+i)z, where 'z' represents a complex number.

Understanding the Problem

The equation involves complex numbers, which are numbers of the form a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit (i² = -1). Our goal is to find the value of 'z' that satisfies the given equation.

Solving the Equation

1. Rearrange the equation:

   • Move all terms with 'z' to one side: (2-3i)z - (1+i)z = 9 - 2i
   • Combine the terms with 'z': [(2-3i) - (1+i)]z = 9 - 2i
   • Simplify: (1-4i)z = 9 - 2i

2. Isolate 'z':

   • Divide both sides of the equation by (1-4i): z = (9-2i) / (1-4i)

3. Rationalize the denominator:

   • Multiply the numerator and denominator by the conjugate of the denominator (1+4i): z = [(9-2i)(1+4i)] / [(1-4i)(1+4i)]
   • Expand the numerator and denominator: z = (9 + 36i - 2i - 8i²) / (1 + 16)
   • Simplify using i² = -1: z = (9 + 34i + 8) / 17
   • Combine real and imaginary terms: z = (17 + 34i) / 17

4. Express in standard form:

   • Simplify the fraction: z = 1 + 2i

Conclusion

Therefore, the solution to the equation (2-3i)z - (9-2i) = (1+i)z is z = 1 + 2i. This process involved manipulating the equation, isolating 'z', rationalizing the denominator, and finally expressing the solution in standard form (a + bi).