Complex Number Question (1 Viewer)

pikachu975 · Oct 7, 2016

Hi, I'm having trouble with this question from Cambridge:

http://prntscr.com/cqwz3l

Any help? All parts, i'm not sure how to do it. Thanks for any responses!

jathu123 · Oct 7, 2016

hopefully this is clearer than before

$$\noindent a) Since all the cooefficients are real, due to the conjugate root theorem, $3+2i$ is also a root. Now the sum of roots is $-b=3+2i+3-2i=6\Rightarrow b=-6\\ $ The product of the roots is $c=(3+2i)(3-2i)=13\\ \\ $b) Since all the coefficients are real, we can use the conjugate root theorem again. If one root is $ Re(\alpha )+2i,$ then the other must be its conjugate which is $Re(\alpha)-2i.$ Doing the sum of roots, $-6=Re(\alpha )+2i+Re(\alpha )-2i =2Re(\alpha ) \Rightarrow Re(\alpha )=-\frac{6}{2}=-3.$ $ \\\therefore$ the roots are $-3+2i $ and $-3-2i$. Now doing the products of the roots, $k=(-3+2i)(-3-2i)=13\\ \\$ c) Since the roots aren't all real, we cannot use the conjugate root theorem. Let the other root be $z$. Doing the sum of roots, $ -\frac{b}{a}=3+i=1-2i+z\\\Rightarrow z=3+i-1+2i=2+3i. $ Now doing the products of roots, \\$k=(1-2i)(2+3i)=8-i$

pikachu975 · Oct 7, 2016

jathu123 said:
hopefully this is clearer than before

$$\noindent a) Since all the cooefficients are real, due to the conjugate root theorem, $3+2i$ is also a root. Now the sum of roots is $-b=3+2i+3-2i=6\Rightarrow b=-6\\ $ The product of the roots is $c=(3+2i)(3-2i)=13\\ \\ $b) Since all the coefficients are real, we can use the conjugate root theorem again. If one root is $ Re(\alpha )+2i,$ then the other must be its conjugate which is $Re(\alpha)-2i.$ Doing the sum of roots, $-6=Re(\alpha )+2i+Re(\alpha )-2i =2Re(\alpha ) \Rightarrow Re(\alpha )=-\frac{6}{2}=-3.$ $ \\\therefore$ the roots are $-3+2i $ and $-3-2i$. Now doing the products of the roots, $k=(-3+2i)(-3-2i)=13\\ \\$ c) Since the roots aren't all real, we cannot use the conjugate root theorem. Let the other root be $z$. Doing the sum of roots, $ -\frac{b}{a}=3+i=1-2i+z\\\Rightarrow z=3+i-1+2i=2+3i. $ Now doing the products of roots, \\$k=(1-2i)(2+3i)=8-i$

Thanks for the help. Does it take long to do latex?

jathu123 · Oct 8, 2016

pikachu975 said:
Thanks for the help. Does it take long to do latex?

No worries! Hope that helped.
And yeah it does take a tiny bit longer than typing it normally

leehuan · Oct 8, 2016

pikachu975 said:
Thanks for the help. Does it take long to do latex?

Not really if you can type fast though (y)

Complex Number Question (1 Viewer)

pikachu975

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pikachu975

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jathu123

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leehuan

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