I. \(1 + 4i > 3 + 2i\)
II. \(2 + 3i < 3 + 4i\)
III. \(4 + 3i > 3 + 4i\)
where \(i = \sqrt{-1}\)
How many of the above are valid ?
Analyzing Complex Number Inequalities Validity
The core principle is that standard ordering operators like \(>\) (greater than) and \(<\) (less than) are not defined for complex numbers if their imaginary parts are non-zero. These operators only apply to real numbers.
Let's examine each inequality based on this principle:
Inequality I: \(1 + 4i > 3 + 2i\)
- Complex numbers involve both a real part and an imaginary part.
- Standard comparison operators (\(>\), \(<\)) are undefined for complex numbers unless they are purely real (imaginary part is zero).
- Therefore, the inequality \(1 + 4i > 3 + 2i\) is not valid.
Inequality II: \(2 + 3i < 3 + 4i\)
- Similar to Inequality I, the comparison operator \(<\) is not applicable to complex numbers \(2 + 3i\) and \(3 + 4i\) as they have non-zero imaginary parts.
- Thus, the inequality \(2 + 3i < 3 + 4i\) is invalid.
Inequality III: \(4 + 3i > 3 + 4i\)
- Again, the operator \(>\) cannot be meaningfully applied to compare the complex numbers \(4 + 3i\) and \(3 + 4i\).
- Hence, the inequality \(4 + 3i > 3 + 4i\) is invalid.
Conclusion on Validity
Since none of the provided inequalities involving complex numbers are mathematically valid according to standard definitions, the correct count is zero.
- Inequality I: Invalid
- Inequality II: Invalid
- Inequality III: Invalid
Therefore, zero out of the three inequalities are valid.
Important Questions from Complex Numbers
- If $\omega$ is a complex cube root of unity, then the value of $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)$ is:
If A + iB = tan (x + iy), then the value of tan 2x is?
The value of \({\left( {\frac{{\cos \theta + i\sin \theta }}{{i\cos \theta + \sin \theta }}} \right)^4}\) is:
The smallest positive integer n for which \(\left(\dfrac{1+i}{1-i}\right)^n=1\) , is
If ω is cube root of unity, then (3 + ω + 3ω 2) 6 is equal to
