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How do you multiply imaginary numbers? - Answers
First, let's make sure we are not confusing imaginary numbers with complex numbers. Imaginary (sometimes called "pure imaginary" for clarity) numbers are numbers of the form ai, where a is a real number and i is the principal square root of -1. To multiply two imaginary numbers ai and bi, start by pretending that i is a variable (like x). So ai x bi = abi2. But since i is the square root of -1, i2=-1. So abi2=-ab. For example, 6i x 7i =-42. 5i x 2i =-10. (-5i) x 2i =-(-10)= 10. Complex numbers are numbers of the form a+bi, where a and b are real numbers. a is the real part, bi is the imaginary part. To multiply two complex numbers, again, just treat i as if it were a variable and then in the final answer, substitute -1 wherever you see i2. Hence (a+bi)(c+di) = ac + adi + bci + dbi2 which simplifies to ac-db + (ad+bc)i. For example: (2+3i)(4+5i) = 8 + 10i +12i + 15i2= 8 + 10i + 12i - 15 = -7 + 22i
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How do you multiply imaginary numbers? - Answers
First, let's make sure we are not confusing imaginary numbers with complex numbers. Imaginary (sometimes called "pure imaginary" for clarity) numbers are numbers of the form ai, where a is a real number and i is the principal square root of -1. To multiply two imaginary numbers ai and bi, start by pretending that i is a variable (like x). So ai x bi = abi2. But since i is the square root of -1, i2=-1. So abi2=-ab. For example, 6i x 7i =-42. 5i x 2i =-10. (-5i) x 2i =-(-10)= 10. Complex numbers are numbers of the form a+bi, where a and b are real numbers. a is the real part, bi is the imaginary part. To multiply two complex numbers, again, just treat i as if it were a variable and then in the final answer, substitute -1 wherever you see i2. Hence (a+bi)(c+di) = ac + adi + bci + dbi2 which simplifies to ac-db + (ad+bc)i. For example: (2+3i)(4+5i) = 8 + 10i +12i + 15i2= 8 + 10i + 12i - 15 = -7 + 22i
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How do you multiply imaginary numbers? - Answers
First, let's make sure we are not confusing imaginary numbers with complex numbers. Imaginary (sometimes called "pure imaginary" for clarity) numbers are numbers of the form ai, where a is a real number and i is the principal square root of -1. To multiply two imaginary numbers ai and bi, start by pretending that i is a variable (like x). So ai x bi = abi2. But since i is the square root of -1, i2=-1. So abi2=-ab. For example, 6i x 7i =-42. 5i x 2i =-10. (-5i) x 2i =-(-10)= 10. Complex numbers are numbers of the form a+bi, where a and b are real numbers. a is the real part, bi is the imaginary part. To multiply two complex numbers, again, just treat i as if it were a variable and then in the final answer, substitute -1 wherever you see i2. Hence (a+bi)(c+di) = ac + adi + bci + dbi2 which simplifies to ac-db + (ad+bc)i. For example: (2+3i)(4+5i) = 8 + 10i +12i + 15i2= 8 + 10i + 12i - 15 = -7 + 22i
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- og:descriptionFirst, let's make sure we are not confusing imaginary numbers with complex numbers. Imaginary (sometimes called "pure imaginary" for clarity) numbers are numbers of the form ai, where a is a real number and i is the principal square root of -1. To multiply two imaginary numbers ai and bi, start by pretending that i is a variable (like x). So ai x bi = abi2. But since i is the square root of -1, i2=-1. So abi2=-ab. For example, 6i x 7i =-42. 5i x 2i =-10. (-5i) x 2i =-(-10)= 10. Complex numbers are numbers of the form a+bi, where a and b are real numbers. a is the real part, bi is the imaginary part. To multiply two complex numbers, again, just treat i as if it were a variable and then in the final answer, substitute -1 wherever you see i2. Hence (a+bi)(c+di) = ac + adi + bci + dbi2 which simplifies to ac-db + (ad+bc)i. For example: (2+3i)(4+5i) = 8 + 10i +12i + 15i2= 8 + 10i + 12i - 15 = -7 + 22i
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