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# (3-i)+(1-2i)= solve

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For arithmetic with complex numbers, treat the i as a named variable, and keep in mind i * i = -1 for multiplicative steps (not needed for this special case). I'll demonstrate in small steps, more than I'd do normally:

(3 - i) + (1 - 2i)
= 3 - i + 1 -2i [associative law]
= 3 + 1 - i - 2i [commutative law]
= ( 3 + 1 ) + ( -i -2i ) [associative law]
= 4 + (-3i) [consolidated]
= 4 - 3i [consolidated]

You could do the same by using component notation for the complex numbers, avoiding the "i":

(3 - i ) + (1 - 2i )
= (3; -1) + (1; -2) [component notation (real; imaginary)]
= (4; -3) [addition by components]
= 4 - 3i [i-notation]

Posted on Mar 20, 2014

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Posted on Jan 02, 2017

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