(2+3i/3+4i)(2-3i/3-4i) Is Real

Jasonbradley

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Jun 16, 2024

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Proving that (2+3i/3+4i)(2-3i/3-4i) is Real

This article will demonstrate why the expression (2+3i/3+4i)(2-3i/3-4i) results in a real number.

Understanding Complex Numbers

Before diving into the calculation, let's refresh our understanding of complex numbers.

A complex number is a number of the form a + bi, where a and b are real numbers, and i is the imaginary unit, defined as the square root of -1. The a part is called the real part and the bi part is called the imaginary part of the complex number.

Calculation and Simplification

Let's simplify the expression step by step:

1. Simplify the fractions:

   • (2 + 3i)/(3 + 4i) can be simplified by multiplying both numerator and denominator by the conjugate of the denominator, which is (3 - 4i).

     (2 + 3i)(3 - 4i) / (3 + 4i)(3 - 4i) = (6 + 9i - 8i + 12) / (9 + 16) = (18 + i) / 25

   • Similarly, (2 - 3i)/(3 - 4i) can be simplified by multiplying both numerator and denominator by (3 + 4i):

     (2 - 3i)(3 + 4i) / (3 - 4i)(3 + 4i) = (6 - 9i + 8i + 12) / (9 + 16) = (18 - i) / 25

2. Multiply the simplified fractions:

   [(18 + i) / 25] * [(18 - i) / 25] = (18 + i)(18 - i) / 625

3. Expand the numerator:

   (18 + i)(18 - i) = 18² - i² = 324 + 1 = 325

4. Final result:

   325 / 625 = 1 / 5

Conclusion

The expression (2+3i/3+4i)(2-3i/3-4i) simplifies to 1/5, which is a real number. This illustrates a key property of complex numbers: multiplying a complex number by its conjugate results in a real number.