Let u = ( 1-i,2i ), v = ( 1+i,-2 ), w = ( 2,-2+2i ) be vectors in the
complex vector space.
(a). Evaluate (3+i)u.
I'll go through all the steps. I always put constants in parentheses, with
no space after the 1st parenthesis or before the 2nd parenthesis. I put
vectors in parentheses too, but with a space after the first parenthesis and
before the 2nd parenthesis, and, of course, a comma between the components.
(3+i)( 1-i,2i ) = ( (3+i)(1-i),(3+i)(2i) ) = ( 3-3i+i-i²,6i+2i² ) =
( 3-2i-(-1),6i+2(-1) ) = ( 3-2i+1,6i-2 ) = ( 4-2i,-2+6i )
(b). (1+i)v.
(1+i)( 1+i,-2 ) = ( 2i,-2-2i ) <-- didn't go through the steps here.
(c). Determine of possible a complex scalar c such that v=cu
v = cu, let c = a+bi
( 1+i,-2 ) = (a+bi)( 1-i,2i )
( 1+i,-2 ) = ( (a+b)+(b-a)i,-2b+2ai )
We equate 1st components:
1+i = (a+b)+(b-a)i
Solving that system gives a=0, b=1, c=(0+1i)=(i)
Now we check to see if the 2nd coordinates are also equal
using a=0, b=1, c= (0+i) = (i)
-2 =?= -2b+2ai
-2 =?= -2(1)+2(0)i
-2 =?= -2
Yes they are, so a complex scalar c is possible, c = (0+1i) = (i)
Edwin