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The Strange Behavior of i (Posted on 2023-06-26) Difficulty: 1 of 5
Let us consider a positive real number A.
For the imaginary number i, we know that: i*i = -1.
Therefore, i*i*i*i = (-1)*(-1)= 1
Hence, A*i*i*i*i = A
or √(A*i*i*i*i) = √A
or, √(A)*(i*i) = √A,
since sqrt(i*i*i*i) = i*i
or, - √(A) = √(A)

Provide a valid reason for this apparent inconsistency.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
soln Comment 1 of 1
The strict result for using the radical symbol as a principle square root is:
 R exp(i theta), with R a positive real and (-pi .lt. theta .le. pi) 

For  sqrt(i*i * i*i), theta is pi/2, not 3 pi/2, 
so sqrt(i*i * i*i) = -1 is false. 


Edited on June 27, 2023, 2:45 am
  Posted by Steven Lord on 2023-06-26 09:47:35

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