[ 2 - 3i ] is.
-2 - 3i
- 2 - 3i
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
This is a Bronsted question. Hs- is the acid in this which makes H2O a base. Therefore S-2 is the conjugate base and the H3O+ hydronium ion is the conjugate acid.
The complex conjugate of 2-3i is 2+3i.
-2 - 3i
- 2 - 3i
The conjugate of 2 + 3i is 2 - 3i, and the conjugate of 2 - 5i is 2 + 5i.
-2 + 3iThe additive inverse: -(-2 + 3i) = 2 - 3iThe conjugate: -2 - 3i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
The product is a^2 + b^2.
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.
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
The conjugate of a complex number is formed by changing the sign of its imaginary part. Since (6 + \sqrt{2}) is a real number (with no imaginary part), its conjugate is simply itself: (6 + \sqrt{2}).
(1+i)3 = 1 + 3i - 3 - i = -2 + 2i This is a complex number, and therefore cannot be plotted on a Cartesian plane.