document.write( "Question 6614: How do you simplify ((1-2i)/(3+i)) and write in a+bi form? \n" ); document.write( "

Algebra.Com's Answer #3596 by prince_abubu(198) $\"\"$ $\"About$

You can put this solution on YOUR website!
You believe that anything divided by itself = 1, right? And multiplying something by 1 doesn't change the original value?\r
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\n" ); document.write( "\n" ); document.write( "We'll multiply both numerator and denominator with the denominator's conjugate. The conjugate is exactly the complex number a +/- bi EXCEPT that you flip the sign before the bi term. Since you've got 3 + i, the conjugate would be 3 - i. We'll multiply both numerator and denominator by the conjugate.\r
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\n" ); document.write( "\n" ); document.write( " $\"+%28%281+-+2i%29%2F%283%2Bi%29%29%2A%28%283-i%29%2F%283-i%29%29+\"$ really is multiplying your original rational expression by 1, so it doesn't change anything. This is just a trick you can use.\r
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\n" ); document.write( "\n" ); document.write( "Let's take care of the numerator first by performing FOIL:\r
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\n" ); document.write( "\n" ); document.write( " $\"+%281+-+2i%29%283+-+i%29+=+3+-+i+-+6i+%2B+2i%5E2+=+3+-+7i+-+2+=+1+-+7i+\"$ \r
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\n" ); document.write( "\n" ); document.write( "Let's take the denominator and do FOIL\r
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\n" ); document.write( "\n" ); document.write( " $\"+%283+%2B+i%29%283+-+i%29+=+9+-+3i+%2B+3i+-+%28-1%29+=+10+\"$ \r
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\n" ); document.write( "\n" ); document.write( "So, numerator over denominator would give you $\"+%281+-+7i%29%2F10+\"$ . They want it in a + bi form. Remember your fraction rule that says $\"+%28a+-+b%29%2Fc+=+a%2Fc+-+a%2Fb+\"$ . So in this case, your answer will be $\"+1%2F10+-+%287%2F10%29i+\"$ \n" ); document.write( "

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