Complex number simplification

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10 Replies
iTeachChem Helper
@Apu
iTeachChem Helper
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.
Nimboi [ping if answering]
bro what are they doing here
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Nimboi [ping if answering]
what is that
SirLancelotDuLac
Let there be a polynomial with one root as -5+4i. For coeffecients being real its conjugate must also be a root of this polynomial. In line 2 we find the said polynomial. Now, let q(x)=p(x).r(x)+g(x) where p(x) is the polynomial we found in 2 and q(x) is the biquadratic we have. Plugging in the complex number in q(x), the p(x).r(x) term becomes zero and the remainder, g(x) gives the requisite.
Nimboi [ping if answering]
what is this step why'd we do that
SirLancelotDuLac
Remainder theorem kinda stuff.
Nimboi [ping if answering]
ouh i completely forgot that existed
SirLancelotDuLac
Like jabh degree greater than 2 hoti hai, this is an approach.
Nimboi [ping if answering]
uhh oke well ill do remainder theorem and some questions then see if i get this better ill close it then thank u

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