| 3-i - 1/(3-i)| = | 3-i - (3+i)/10|= | (27-11i)/10| = sqrt(17/2)
1/(1+ 3i)
3i where i is the square root of negative one.3i x 3i = -9
4
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 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.
1/(1+ 3i)
3i where i is the square root of negative one.3i x 3i = -9
(3i)5 = 35 x i5 = 243 x i2 x i2 x i = 243 x (-1) x (-1) x i = 243i
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
The given term is 3√-1. Since √-1 = i, we can rewrite the term as 3i.
x2 + 9 has no real factors. Its complex factors are (x + 3i) and (x - 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
2
3sqrt(- 1 ) = 3i --------
4
3sqrt(- 1 ) = 3i --------
Not complex numbers (such as 4+3i), the square of these are negative because i is defined as the square root of -1