What is the domain when a function has an imaginary number?

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

7 0

The imaginary numbers are often presented on a real-imaginary graph where the imaginary numbers are on the vertical axis.

A function that has imaginary numbers would have a domain that consists a range of numbers on the horizontal axis. Since the horizontal axis represents real numbers, hence the domain of the function (providing there's no limitation) is all the numbers on the horizontal axis

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It’s been a while since I did Math but, in most scenarios, when you solve for an equation with only 1 x variable and you substit

patriot [66]

Let's replace every copy of x with 7 and simplify.

$\sqrt{5x+1}+9 = 3\\\\\sqrt{5*7+1}+9 = 3\\\\\sqrt{35+1}+9 = 3\\\\\sqrt{36}+9 = 3\\\\6+9 = 3 \ \text{ .... note the 6 on the left isn't negative}\\\\15 = 3\\\\$

Which is clearly false. This is probably the set of steps you followed to get "no solution".

Why is there no solution? Well let's subtract 9 from both sides to isolate the square root

$\sqrt{5x+1}+9 = 3\\\\\sqrt{5x+1}+9-9 = 3-9\\\\\sqrt{5x+1} = -6\\\\$

Recall that the result of a square root operation is never negative. The range of $y = \sqrt{x}$ and $y = \sqrt{5x+1}$ is the set of nonnegative numbers. There is no way we can have the left hand side result in -6