Just multiply everything out you end up with b=4a so the solution set is (a, b) = (k, 4k) for any non-zero k; Non zero because of domain issues, a+bi ≠ 0, k + 4k ≠ 0, 5k ≠ 0, k ≠ 0
I used the definition of "proportion", and consequently I applied the property according to which the product of the means is equal to the product of the extremes, arriving at the same result.
The set of solutions is z = r(1+4i) where r is any nonzero real number.
Just multiply everything out you end up with b=4a so the solution set is (a, b) = (k, 4k) for any non-zero k;
Non zero because of domain issues, a+bi ≠ 0, k + 4k ≠ 0, 5k ≠ 0, k ≠ 0
I used the definition of "proportion", and consequently I applied the property according to which the product of the means is equal to the product of the extremes, arriving at the same result.
Seeing that i(z-bar)/z is a constant means that, if w is a solution, then kw is also a solution, because the k comes out of the conjugate and cancels.
Nice
Reduce to a single variable you dividing top and bottom by a. Then put k=b/a and the answer is obvious.
*a=k, b=4k (k là số thực khác 0)*
k must be nonzero.
😊🎉😊👍👍👍😊🎉😊
By looking at the problem, one can see that a=1 and b=4.
Yes... but there are other solutions 😉
k € R* and not real