document.write( "Question 1141474: Let u = (1-i,2i), v = (1+i,-2), w = (2,-2+2i) be vectors in the complex vector
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Algebra.Com's Answer #762491 by Edwin McCravy(20055) $\"\"$ $\"About$

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

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