First HP CASE - implicit multiplication of complex numbers

When you make the implicit multiplication of complex numbers in CASE the first mode (6975) does the following:

Enter: (2 + i)(2-i)
Rewritten (wrongly) as: 2 + i (2-i)
Returns: 2 + I

This happens in two (a, b) and (a + b * i) formats:

(2.1) (2, - 1) becomes 2,1(2,-1) and returns the value 1

Explicit multiplication of complex numbers seems to work correctly:

Enter: (2 + i) *(2-i)
Returns: 5

Sometimes, the rewriting of implicit multiplication treats the imaginary part of the first complex number as a function call with the second number complex as parameters to the function.

Apparently this tangent actually proved useful.  He looked more closely into the use of parentheses.

Review: (1 + i)(1-i)

What seems to take place the second term, (1 - i), apparently ignored is that CASE trying to make a substitution as in the following example:

(1 + x) (a + b) becomes (1 + (a + b)) due to a substitution of (a + b), the symbol 'x '.

However, if this substitution occurs when a value is assigned to 'x', the substitution fails and the initial value of 'x' remains.  This creates the illusion that (a + b) is ignored.

x: = 2

(1 + x) (a + b) becomes 1 + x and the result is 3 because 'x' remained the value 2.

I think that 'I' is treated as if 'i' was assigned to the value sqrt(-1).  The substitution is not in the same way that when a value is assigned to 'x '.

Remember to purge (x).

Tags: HP Tablets

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