document.write( "Question 179579: 4-2i ÷ 3+i\r
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\n" ); document.write( "\n" ); document.write( "-5-3i ÷ 2-2i
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Algebra.Com's Answer #134547 by Mathtut(3670)\"\" \"About 
You can put this solution on YOUR website!
\"%284-2i%29%283-i%29+%2F+%283%2Bi%29%283-i%29\"
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\n" ); document.write( "\"%2812-4i-6i%2B2i%5E2%29%2F%289%2B1%29\"
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\n" ); document.write( "\"%2812-10i-2%29%2F10\"
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\n" ); document.write( "\"%2810-10i%29%2F10\"
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\n" ); document.write( "\"highlight%281-i%29\"
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\n" ); document.write( "\"%28-5-3i%29%282%2B2i%29+%2F+%282-2i%29%282%2B2i%29\"
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\n" ); document.write( "\"%28-10-10i-6i-6i%5E2%29%2F%284%2B4%29\"
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\n" ); document.write( "\"%28-4-16i%29%2F8\"
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\n" ); document.write( "\"highlight%28%28-1%2F2%29-2i%29\"\r
\n" ); document.write( "\n" ); document.write( "I will do one. and get you started on the other.. you must multiply by the conjugate to get rid of the complex number in the denominator
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\n" ); document.write( "1.3+i√5/3-i√5
\n" ); document.write( ":so we have \"%281.3%2Bi%2Asqrt%285%29%29%2F%283-i%2Asqrt%285%29%29\"
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\n" ); document.write( "so the conjugate of \"3-+sqrt+%285%29\" is \"3+%2Bi%2Asqrt+%285%29\"
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\n" ); document.write( "\"%283.9%2B2.3i%2Asqrt%285%29-5%29%2F%289%2B5%29\"
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\n" ); document.write( "\n" ); document.write( "\"%281.1-2.3i%2Asqrt%285%29%29%2F14\"
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\n" ); document.write( " 2.4-i√2/i√2 the conjugate for this is \"-i%2Asqrt%282%29\" repeat what took place in the problem I just solved...multiplying both top and bottom by the conjugate\r
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