document.write( "Question 1104383: Prove |z1/z2|=|z1|/|z2| without using the polar form. \n" ); document.write( "

Algebra.Com's Answer #719479 by math_helper(2461) $\"\"$ $\"About$

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$\"+abs%28z1%29+=+abs%28a%2Bbi%29+=+sqrt%28a%5E2%2Bb%5E2%29+\"$
\n" ); document.write( " $\"+abs%28z2%29+=+abs%28c%2Bdi%29+=+sqrt%28c%5E2%2Bd%5E2%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "notice that $\"+abs%28z1%29%5E2+=+%28a%2Bbi%29%28a-bi%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "

\n" ); document.write( "= $\"+abs%28+%28%28ac%2Bbd%29+%2B+i%28bc-ad%29%29+%2F+%28c%5E2%2Bd%5E2%29%29+\"$
\n" ); document.write( "=

\n" ); document.write( ": : :
\n" ); document.write( "After multiplying out and canceling a 2abcd with a -2abcd in the numerator...
\n" ); document.write( ": : :
\n" ); document.write( "=

\n" ); document.write( "= $\"+sqrt%28+%28%28a%5E2%2Bb%5E2%29%28c%5E2%2Bd%5E2%29%29%2F%28c%5E2%2Bd%5E2%29%5E2%29+\"$
\n" ); document.write( "= $\"+sqrt+%28%28a%5E2%2Bb%5E2%29%2F%28c%5E2%2Bd%5E2%29%29+\"$ = $\"+sqrt%28a%5E2%2Bb%5E2%29+%2F+sqrt%28c%5E2%2Bd%5E2%29+\"$
\n" ); document.write( "= $\"+abs%28z1%29+\"$ / $\"+abs%28z2%29+\"$
\n" ); document.write( "Done. \n" ); document.write( "

\n" );