What can you say about i^0 where i is imaginary number iota.
Is it equal to 0? If yes then prove it. If anything else, give appropriate reason.
Answers
In such cases, we write √−3 as √−3 = √−1 × √3. This would give the solution of the above quadratic equation to be: x = (−1 ± √3i)/2. Hence, the value of iota is helpful in solving square roots with negative values.
Thus, the value of iota is, i = √−1.
Answer:
i^0=1
Anything with power 0 is always 1.
For example, let us take 2^0. In this case we are not actually multiplying the number 2 by 0.
We define 2^0 = 1, so that each power of 2 is one factor of 2 larger than the last, e.g., 1,2,4,8,16,32...
This involves the rules of exponents particularly division.
If a is a number and x and y are also numbers, then according to the rule of division for powers with the same base,
a^x/a^y = a^(x - y).
It says the quotient of two powers with the same base is equal to the common base raise to the exponent equal to the difference between x and y.
So, if x = y, then a^x/a^y = a^(x -y) = a^0
But a^x is equal to a^y, since x = y; hence a^x/a^y = 1
Therefore, by Transitive property of Equality,
a^0 = 1
Thus, this result says that number raised to the power zero is equal to 1.
Yes, i^0=1