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.

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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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