for example, in \$displaystylefraca+bin+zi\$, you would multiply both by the complex conjugate of the denominator, \$n-zi\$, to get rid of the facility number in the denominator. Wouldn"t multiply both by \$i\$ to obtain \$i^2\$ on the bottom and also top eliminate the complex numbers?  For any facility number \$z\$, multiply by the conjugate always gives a nonnegative actual number:\$\$(a+bi)(a-bi) = a^2+b^2.\$\$While occasionally you can multiply a facility number by some other facility number to obtain a real (e.g., you have the right to multiply a completely imaginary number by \$i\$), the conjugate always works.

You are watching: When do you multiply by the conjugate  Precisely the actual multiples that the conjugate suffice come rationalize a denominator \$ m:zin keolistravelservices.combb C:.:\$ Proof: if \$ m z e0 \$ and also \$ m y:z = r in keolistravelservices.combb R \$ then \$ m: y:z:z': = r:z': \$ so \$ m y = z':r/(z:z') = s:z', s: = r/(z:z')in keolistravelservices.combb R:.:\$ conversely, if \$ m y = s:z':, sinkeolistravelservices.combb R \$ then \$ m y:z = s:z':zin keolistravelservices.combb R:.\$ This is best thought of in terms of the polar representation of a complicated number. Permit \$z = r exp (i heta)\$. What carry out we need to multiply through to turn this into a real number? \$exp (- ns heta)\$, the course. Any type of real many of it additionally works, and we have actually one easily at hand. Provided \$z = x + i y = r exp(i heta)\$, we deserve to write \$r exp(-i heta) = overlinez = x - i y\$.

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