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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.
Bing
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.
DuckDuckGo
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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- og:descriptionI 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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