(1-i)(z+1-2i)-3+2i=0

Jasonbradley

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Jun 16, 2024

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Solving the Complex Equation: (1-i)(z+1-2i)-3+2i=0

This article will guide you through solving the complex equation (1-i)(z+1-2i)-3+2i=0. We'll break down the steps and utilize the properties of complex numbers to find the solution for z.

Understanding Complex Numbers

Before we start, let's review some essential concepts about complex numbers:

• Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit (√-1).
• The imaginary unit i has the property that i² = -1.

Solving the Equation

1. Expand the product: Begin by expanding the product on the left side of the equation:

   (1-i)(z+1-2i) = z + 1 - 2i - iz - i + 2i²

   Remember that i² = -1, so:

   (1-i)(z+1-2i) = z + 1 - 2i - iz - i - 2

   Combining like terms:

   (1-i)(z+1-2i) = (z - iz) + (-2i - i) + (1 - 2)

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

2. Combine the terms: Now, rewrite the equation with all terms combined:

   (z - iz) - 3i - 1 - 3 + 2i = 0

3. Isolate z: Group terms with z and separate them from the constant terms:

   (z - iz) = 3i + 4

4. Factor out z: Factor out z from the left side of the equation:

   z(1 - i) = 3i + 4

5. Solve for z: Divide both sides of the equation by (1-i) to isolate z:

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

6. Rationalize the denominator: To get rid of the imaginary term in the denominator, multiply both numerator and denominator by the complex conjugate of the denominator:

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

   Simplify:

   z = (3i + 4 + 3i² + 4i) / (1² - i²)

   z = (7i + 1) / (1 + 1)

   z = (7i + 1) / 2

7. Express the solution in standard form: Finally, express the solution in the standard form a + bi:

   z = (1/2) + (7/2)i

Conclusion

The solution to the complex equation (1-i)(z+1-2i)-3+2i=0 is z = (1/2) + (7/2)i. By understanding the properties of complex numbers and applying algebraic techniques, we were able to successfully solve the equation.