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**garrisontrantow****Member**- Registered: 2021-01-11
- Posts: 1

Hi,

How to right imaginary number with inequalities and in interval notation like this :

x^2 ≥ -1

it is either x ≥ + √-1

or x ≥ - √-1

so how to right it in number Line? and in interval notation,

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,931

hi garrisontrantow

Welcome to the forum.

I've never encountered an inequality problem involving complex numbers but I don't see why it shouldn't be allowed.

The 'imaginary' access is not called that because it is somehow 'made up' but rather because it is the image of the other axis.

On the real axis > means to the right of, so on the imaginary axis ** I'll define > to mean higher up the line**.

Then I'll divide the exercise into three parts:

(1) If x is a real number. x^2 ≥ 0 for all x, so every real number is a solution.

(2) If x is an imaginary number. You've pretty much solved this already, but consider if x = 2i Then x^2 = -4

-4 is not ≥ -1 using my definition ** because -4 is not higher up the line.

Consider x = 0.5i x^2 = -0.25 This is higher up the line than -1.

And if x -= -0.25i x^2 = -0.25 again. So it looks like the solution is -i ≤ x ≤ +i

(3) If x is a fully complex number. That is to say it has the form a + bi where a and b are both non zero real numbers.

In modulus argument form, let the argument be θ where θ is between 0 and 360 and not equal to 90, 180 and 270.

When you square x the argument is doubled , which means that its argument is also not 0, 90, 180, 270, 360. Thus the square is also a fully complex number. The inequality has no meaning in these circumstances so there are no new solutions here.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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