- #1
dado033
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is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)
dado033 said:is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)
A relation in complex analysis is a set of ordered pairs that represents the correspondence between two sets of complex numbers. It is used to describe the connections between different elements in a complex function or equation.
A relation can be seen as a generalization of a function, where a function is a specific type of relation that assigns exactly one output for each input. In a relation, an input can have multiple corresponding outputs, whereas in a function, each input has only one output.
Relations are important in complex analysis because they allow us to study the behavior and properties of complex functions. They also help us to understand the connections between different points on the complex plane and how they relate to each other.
A relation in complex analysis is considered symmetric if for every input-output pair (a, b) in the relation, there exists an input-output pair (b, a) in the relation as well. In other words, if the order of the input and output can be reversed and the relation still holds, then it is symmetric.
No, a relation cannot be both reflexive and anti-symmetric at the same time. A relation is reflexive if every element is related to itself, and anti-symmetric if no distinct elements are related to each other. These two properties are contradictory, so a relation cannot have both at the same time.