a more detailed solution of an answer
One significant feature of complex numbers is that all polynomial equations of order n, in the complex field, have n solutions. When multiple roots are Given any set of complex numbers {a(0),  … , a(n)}, such that at least one of a(1) to a(n) is non-zero, the equation a(n)*z^n + a(n-1)*z^(n-1) + ... + a(0) has at least one solution in the complex field. This is the Fundamental Theorem of Algebra and establishes the set of Complex numbers as a closed field. [a(0), ... , a(n) should be written with suffices but this browser has decided not to be cooperative!] The above solution is the complex root of the equation. In fact, if the equation is of order n, that is, if the coefficient a(n) is non-zero then, taking account of the multiplicity, the equation has exactly n roots (some of which may be real).
You multiply the numerator and the denominator of the complex fraction by the complex conjugate of the denominator.The complex conjugate of a + bi is a - bi.
You just plug in the coefficients, and do the normal operations. Of course you have to know how to calculate with complex numbers. Assuming the coefficients are real, you may at some moment get the root of a negative number. Say, for instance, you have the square root of minus 2, then the solution of that part is the square root of plus 2, multiplied by i.If the original coefficients are complex, you may have to calculate the root of a complex number. This is a little more complicated. For this, you convert the complex number to polar coordinates - that is, to a length and an angle. Then, to actually take the square root, you take half the angle, and the square root of the distance - and convert back to rectangular coordinates (separating the real and the imaginary part). (For the second solution, add 180 degrees to the angle.)
A complex Query is a quiery that is much more complex than a normal quiery so search up complex then quiry!!!!
Sure. Just write the corresponding solution:"multiplicative inverse of x" = "additive inverse of x"1/x = -xIf you solve this, you get two solutions.
Add salt to the solution. Change the temperature of the solution. Let the solvent evaporate in increase the concentration of the soluble complex. Change the pH of the solution.
Yes, beer is a complex homogeneous solution (unfiltered beer is a heterogeneous solution).
No.
Actually you can. It all depends on whether a complex number is a valid solution in a certain context. In some contexts, complex numbers make sense, in others, they don't.Actually you can. It all depends on whether a complex number is a valid solution in a certain context. In some contexts, complex numbers make sense, in others, they don't.Actually you can. It all depends on whether a complex number is a valid solution in a certain context. In some contexts, complex numbers make sense, in others, they don't.Actually you can. It all depends on whether a complex number is a valid solution in a certain context. In some contexts, complex numbers make sense, in others, they don't.
Gasoline is a complex mixture of hydrocarbons and additives.
benidict's solution
The problem was so complex that it required a team of experts to analyze and find a solution.
Sea water is a complex solution.
Hunger is a very complex issue, so there is no onesolution for it. GM foods can be part of the solution.
Complex retains their identity in the solution form and double salt doesn't retain its identity in solution. A Ali Sudais jan Research Scholar Nuclear Medicine Inorganic Chemistry
Here are a few: 0 = 1 x = x + 1 (subtract "x" on each side, and you get the previous one!) x2 = -1 (if you want real numbers; however, it has two solutions in the complex numbers) ln x = -1 (same as above: no solution in the real numbers, but it has a solution in the complex numbers) ln x = 0 (no solution, neither in the real numbers, nor in the complex numbers) 0x = 5
The complex roots of an equation is any solution to that equation which cannot be expressed in terms of real numbers. For example, the equation 0 = x² + 5 does not have any solution in real numbers. But in complex numbers, it has solutions.