The complex conjugate of a complex number is obtained by changing the sign of its imaginary part. For the complex number ( 3i + 4 ), which can be expressed as ( 4 + 3i ), the complex conjugate is ( 4 - 3i ).
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
To write a polynomial function with real coefficients given the zeros 2, -4, and (1 + 3i), we must also include the conjugate of the complex zero, which is (1 - 3i). The polynomial can be expressed as (f(x) = (x - 2)(x + 4)(x - (1 + 3i))(x - (1 - 3i))). Simplifying the complex roots, we have ((x - (1 + 3i))(x - (1 - 3i)) = (x - 1)^2 + 9). Thus, the polynomial in standard form is: [ f(x) = (x - 2)(x + 4)((x - 1)^2 + 9). ] Expanding this gives the polynomial (f(x) = (x - 2)(x + 4)(x^2 - 2x + 10)), which can be further simplified to the standard form.
To simplify the expression ( (7 - 3i) + (4 + 8i) ), combine the real parts and the imaginary parts separately. The real parts are ( 7 + 4 = 11 ), and the imaginary parts are ( -3i + 8i = 5i ). Therefore, the answer is ( 11 + 5i ).
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.
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
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
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.
To simplify the expression ( (7 - 3i) + (4 + 8i) ), combine the real parts and the imaginary parts separately. The real parts are ( 7 + 4 = 11 ), and the imaginary parts are ( -3i + 8i = 5i ). Therefore, the answer is ( 11 + 5i ).
The multiplicative inverse of a complex number is the reciprocal of that number. To find the multiplicative inverse of 4 + i, we first need to find the conjugate of 4 + i, which is 4 - i. The product of a complex number and its conjugate is always a real number. Therefore, the multiplicative inverse of 4 + i is (4 - i) / ((4 + i)(4 - i)) = (4 - i) / (16 + 1) = (4 - i) / 17.
-4-3i
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.
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.