thanks Mercedes Orús-Lacort for your nice answer.

A5. It is important to specify the rules

Answer 5, posted on April 6, 2019

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When working with complex exponents,

it is important to specify which rules are still valid,

which rules are not.

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For example, if we ignored this warning,

we would obtain

1 = sqrt(1) = sqrt[(-1)(-1)]

= [sqrt(-1)] [sqrt(-1)] = [i][i] = i^2 = -1,

which would imply that 1 = -1.

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A second counterexample is the following:

1 = sqrt(1) = 1^(1/2) = [e^(i2π)]^(1/2)

= e^[(i2π)/2] = e^(iπ) = -1.

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With best regards, Jean-Claude

Thank you Dr.Jean-Claude Evard and Dr. Mercedes Orus-Lacort for this useful discussion.

Yes, we must pay attention to the sub-conditions of mathematical laws.

Best regards.

A8. Some of the rules that are no longer valid

Answer 8, posted on April 7, 2019

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The following part of the following Web page

of the free online encyclopedia Wikipedia

contains some of the rules for real numbers

that are no longer valid for complex numbers:

Exponentiation:

https://en.wikipedia.org/wiki/Exponentiation

Failure of power and logarithm identities:

https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities

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With best regards, Jean-Claude

Thanks Dr.Jean-Claude for these helpful links.

You are right, Dr. Mercedes.

Dear Adiya K. Hussein ,

Unfortunately, the previous answers are not complete.

And the complex number (i)5i has multiple values.

And the complete answer is

we know that log(i)=log(e^{i((π/2)+2kπ)})= i((π/2)+2kπ), k∈Z.

Using the definition of complex power.

(i)5i = e5ilog(i) = e^5i(i((π/2)+2kπ))= e^-5((π/2)+2kπ)), k∈Z.

As a special case : the principal value is at k = 0.

the principal value of (i)5i = e^{-5((π/2))}.

Best regards

Thank you Professor Issam, for this important clarification.