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To find the complex conjugate of a number, change the sign in front of the imaginary part. Thus, the complex conjugate of 14 + 12i is simply 14 - 12i.

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Q: Find the complex conjugate of 14 plus 12i?
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What is conjugate of 4i open bracket -2 -3i close bracket?

4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i


What is the answer to this -5-12i?

This is a complex number, not an algebraic expression. The letter i represents the imaginary unit (which is equal to sqrt(-1)). Graphiclly, with real numbers on a horizontal axis, and imaginary numbers on a vertical axis, this means starting at the origin, go to the left 5 units, and then go down 12 units.


Evaluate and express this complex number in standard form the absulote value of 5-12i?

The absolute value of a complex number is it's magnitude (distance from the origin). Think about complex numbers graphically, with reals on the horizontal axis, and imaginaries on the vertical axis. Now you have a right triangle: From the origin move to the right 5 units, then move down 12 units. The absolute value, or magnitude, is the length of the hypotenuse. For this triangle, it is 13: sqrt(5^2 + 12^2) = sqrt(25+144) = sqrt(169) = 13. For magnitudes, we are only interested in the positive square root.


How do you multiply imaginary numbers?

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