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https://math.answers.com/basic-math/How_do_you_find_LCD_in_complex_numbers

How do you find LCD in complex numbers? - Answers

I don't know if this will result in a least common denominator or not, but here is a system that you can use. Suppose you have the complex fraction: (a + bi)/(c + di) where {a,b,c & d} are all real numbers, and i is the imaginary unit number. What I would do is get the denominator to a real number, then use this same procedure for other fractions that you want to add or subtract, then find the LCD between those fractions will real-number denominators.So the first step is to get the fraction to have a real-number denominator. Do this by multiplying the numerator and denominator by the complex conjugate of the denominator. The conjugate of (c + di) is (c - di), so (c + di)*(c - di) = c² - cdi + cdi + d² = c² + d², and (a + bi)*(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i,so the new equivalent fraction equals:(ac - bd) + (ad + bc)i----------------------c² + d²Now do the same for the other fractions. You will have fractions in which all denominators are real numbers, then you can find LCD between these new equivalent fractions.



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How do you find LCD in complex numbers? - Answers

https://math.answers.com/basic-math/How_do_you_find_LCD_in_complex_numbers

I don't know if this will result in a least common denominator or not, but here is a system that you can use. Suppose you have the complex fraction: (a + bi)/(c + di) where {a,b,c & d} are all real numbers, and i is the imaginary unit number. What I would do is get the denominator to a real number, then use this same procedure for other fractions that you want to add or subtract, then find the LCD between those fractions will real-number denominators.So the first step is to get the fraction to have a real-number denominator. Do this by multiplying the numerator and denominator by the complex conjugate of the denominator. The conjugate of (c + di) is (c - di), so (c + di)*(c - di) = c² - cdi + cdi + d² = c² + d², and (a + bi)*(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i,so the new equivalent fraction equals:(ac - bd) + (ad + bc)i----------------------c² + d²Now do the same for the other fractions. You will have fractions in which all denominators are real numbers, then you can find LCD between these new equivalent fractions.



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https://math.answers.com/basic-math/How_do_you_find_LCD_in_complex_numbers

How do you find LCD in complex numbers? - Answers

I don't know if this will result in a least common denominator or not, but here is a system that you can use. Suppose you have the complex fraction: (a + bi)/(c + di) where {a,b,c & d} are all real numbers, and i is the imaginary unit number. What I would do is get the denominator to a real number, then use this same procedure for other fractions that you want to add or subtract, then find the LCD between those fractions will real-number denominators.So the first step is to get the fraction to have a real-number denominator. Do this by multiplying the numerator and denominator by the complex conjugate of the denominator. The conjugate of (c + di) is (c - di), so (c + di)*(c - di) = c² - cdi + cdi + d² = c² + d², and (a + bi)*(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i,so the new equivalent fraction equals:(ac - bd) + (ad + bc)i----------------------c² + d²Now do the same for the other fractions. You will have fractions in which all denominators are real numbers, then you can find LCD between these new equivalent fractions.

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      I don't know if this will result in a least common denominator or not, but here is a system that you can use. Suppose you have the complex fraction: (a + bi)/(c + di) where {a,b,c & d} are all real numbers, and i is the imaginary unit number. What I would do is get the denominator to a real number, then use this same procedure for other fractions that you want to add or subtract, then find the LCD between those fractions will real-number denominators.So the first step is to get the fraction to have a real-number denominator. Do this by multiplying the numerator and denominator by the complex conjugate of the denominator. The conjugate of (c + di) is (c - di), so (c + di)*(c - di) = c² - cdi + cdi + d² = c² + d², and (a + bi)*(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i,so the new equivalent fraction equals:(ac - bd) + (ad + bc)i----------------------c² + d²Now do the same for the other fractions. You will have fractions in which all denominators are real numbers, then you can find LCD between these new equivalent fractions.
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