What is complex conjugate of #i#?
2 Answers
Explanation:
Conjugate of any complex number
It depends...
Explanation:
A conjugate of a number is a number that goes with it in the sense that multiplying the two numbers together yields a simpler kind of number.
For numbers involving radicals, we want to find a multiplier that results in a rational result. For complex numbers we want to find a multiplier that lead to a real result.
If
This is popularly known as "the" complex conjugate of
#bar(a+bi) = a-bi#
We find:
#(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2 in RR#
Using this convention, we find:
#bar(i) = bar(0+1i) = 0-1i = -i#
However, please note that this is not the only possible choice of conjugate.
For example:
#i * i = -1 in RR#
So we could call
In fact we could choose any pure imaginary number to use as a conjugate (in the multiplicative sense) for
Why do I stress this ambiguity?
Consider
What would you say is "the" radical conjugate of
We could mechanically choose
#(sqrt(2)-sqrt(3))(sqrt(2)+sqrt(3)) = 2-3 = -1 in QQ#
So
#(sqrt(3)-sqrt(2))(sqrt(2)+sqrt(3)) = 3-2 = 1#
Going back to complex conjugates, the standard complex conjugate
So in general usage, it is conventional to stay with the standard complex conjugate and refer to "complex conjugate pairs" (for example).