# Complex Numbers Question... (1 Viewer)

#### ohexploitable

##### hey
$\bg_white \\\text{i. Complex roots occur as complex conjugates (don't think that's legit since the question says "show").}\\\text{ii. }(x-\alpha)(x-\overline{\alpha})=x^2-2\text{Re}(\alpha)+|\alpha|^2=x^2-2ax+r^2$

#### ohexploitable

##### hey
$\bg_white \\\text{iii. }2(\alpha+\overline{\alpha})=8\text{ (sum of roots)}\\\alpha+\overline{\alpha}=4\text{..........(1)}\\\\\(\alpha\overline{\alpha})^2=49\text{ (product of roots)}\\\alpha\overline{\alpha}=7\text{..........(2)}\\\\\text{solving (1) and (2) simultaneously,}\\\alpha(4-\alpha)=7\\\alpha=2\pm i\sqrt3\\\\P(x)=(x-2+i\sqrt3)^2(x-2+i\sqrt3)^2$

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#### NewiJapper

##### Active Member
Wow! How the hell do you manage to do that doing General Maths!?

#### NewiJapper

##### Active Member
...................lol

#### ohexploitable

##### hey
Wow! How the hell do you manage to do that doing General Maths!?
We do that stuff in General

#### Bored Of Fail 2

##### Banned
cant you do the first part, showing it instead of stating it, you are a mad noob ohexploitable, that parts easy

see below....

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#### deterministic

##### Member
(a) Let P(x)=px^4+qx^3+rx^2+sx+t where p, q, r,s, t are real
P(a)=0 implies pa^4+qa^3+ra^2+sa+t = 0
Show P(conj(a))=0
Sub into P(x), use the fact conj(a^n)=conj(a)^n, pconj(a)=conj(pa) if p is real.
try it yourself