Q: What is the complex conjugate 3i plus 4?

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Use the rules of division for complex numbers. Just divide 1 / (4 + 3i). This requires multiplying numerator and denominator of this fraction by (4 - 3i), to get a real number in the denominator.

this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4

The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.

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To divide by a complex number, write it as a fraction and then multiply the numerator and denominator by the complex conjugate of the denominator - this is formed by changing the sign of the imaginary bit of the number; when a complex number (a + bi) is multiplied by its complex conjugate the result is the real number a² + b² which can be divided into the complex number of the numerator: (-4 - 3i) ÷ (4 + i) = (-4 - 3i)/(4 + i) = ( (-4 - 3i)×(4 - i) ) / ( (4 + i)×(4 - i) ) = (-16 + 4i - 12i + 3i²) / (4² + 1²) = (-16 - 8i - 3) / (16 + 1) = (-19 - 8i)/17

0+3i has a complex conjugate of 0-3i thus when you multiply them together (0+3i)(0-3i)= 0-9i2 i2= -1 0--9 = 0+9 =9 conjugates are used to eliminate the imaginary parts

Use the rules of division for complex numbers. Just divide 1 / (4 + 3i). This requires multiplying numerator and denominator of this fraction by (4 - 3i), to get a real number in the denominator.

To get the complex conjugate, change the sign in front of the imaginary part. Thus, the complex conjugate of -4 + 5i is -4 - 5i.

this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4

There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.

4/3 -2i

To multiply complex numbers you can use the same FOIL rule that you use for multiplying binomials (First, Inside, Outside, Last).(4 - 3i)(5 + 2i) = (4)(5) +(4)(2i) - (3i)(5) - (3i)(2i) = 20 + 8i-15i - 6(i)^2= 20 -7i - 6(-1) = 20 + 6 -7i = 26 -7i.

-4-3i

the magnitude of a complex number is the square root of the sum of the squares of the real and imaginary parts. so......... (4+9)^.5 = ~3.6 and also (9+4)^.5 = ~3.6 so they are both the same.

The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.

The conjugate base of NH4+ is NH3 (ammonia). A conjugate base is formed by the removal of a proton (H+) from the parent acid or cation.