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4 replies

StuartBruff23-Emerald V
23-Emerald V
November 9, 2019

Sneaky, Valery!

 

A complex modulus operator, as opposed to real square root operator, should sort it out.

 

Stuart

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    ptc-57370661-Visitor
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    January 2, 2020
     

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    Pythagoras theorem is valid when using component modulus only.

    When using geometric rules it is absurd to take consideration of -1 mulltiplicant (i^2) /Avdo

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    ttokoro21-Topaz I
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