factorising over complex numbers:

**earthboy** · February 9th 2013, 08:15 PM

Factorize: $3iu^2+26iu+4u^2+3i-4$ over complex numbers, where $i$ represents the imaginary unit.

Thanks

**Prove It** · February 9th 2013, 08:52 PM

$\displaystyle \begin{align*} 3i\,u^2 + 26i\,u + 4u^2 + 3i - 4 &= \left( 4 + 3i \right) u^2 + 26i\,u + 3i - 4 \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{26i}{4 + 3i} \right) u + \left( \frac{3i - 4}{4 + 3i} \right) \right] \\ &= \left( 4 + 3i \right) \left\{ u^2 + \left[ \frac{ 26i \left( 4 - 3i \right) }{\left( 4 + 3i \right) \left( 4 - 3i \right)} \right] u + \left[ \frac{\left( 3i - 4 \right) \left( 4 - 3i \right)}{\left( 4 + 3i \right) \left( 4 - 3i \right) } \right] \right\} \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{104i - 78i^2 }{25} \right) u + \left( \frac{12i - 3i^2 - 16 + 12i}{25} \right) \right] \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{78 + 104i}{25} \right) u + \left( \frac{-13 + 24i}{25} \right) \right] \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{78 + 104i}{25} \right) u + \left( \frac{39 + 52i}{25} \right) ^2 - \left( \frac{39 + 52i}{25} \right) ^2 + \left( \frac{-13 + 24i}{25} \right) \right] \end{align*}$

$\displaystyle \begin{align*} &= \left( 4 + 3i \right) \left[ \left( u + \frac{39 + 52i}{25} \right) ^2 - \left( \frac{ -1183 + 4056i }{625} \right) + \left( \frac{-325 + 600i}{625} \right) \right] \\ &= \left( 4 + 3i \right) \left[ \left( u + \frac{39 + 52i}{25} \right) ^2 - \left( \frac{-858 + 3456i}{625} \right) \right] \\ &= \left( 4 + 3i \right) \left[ \left( u + \frac{ 39 + 52i }{ 25 } \right) ^2 - \left( \frac{\sqrt{ -858 + 3456i }}{25} \right) ^2 \right] \\ &= \left( 4 + 3i \right) \left( u + \frac{39 + 52i - \sqrt{ -858 + 3456i } }{25} \right) \left( u + \frac{ 39 + 52i + \sqrt{ -858 + 3456i } }{ 25 } \right) \end{align*}$

If you want you CAN get a proper complex number for this square root, but it will be in terms of sines and cosines.

**earthboy** · February 9th 2013, 09:14 PM

Originally Posted by Prove It

$\displaystyle \begin{align*} 3i\,u^2 + 26i\,u + 4u^2 + 3i - 4 &= \left( 4 + 3i \right) u^2 + 26i\,u + 3i - 4 \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{26i}{4 + 3i} \right) u + \left( \frac{3i - 4}{4 + 3i} \right) \right] \\ &= \left( 4 + 3i \right) \left\{ u^2 + \left[ \frac{ 26i \left( 4 - 3i \right) }{\left( 4 + 3i \right) \left( 4 - 3i \right)} \right] u + \left[ \frac{\left( 3i - 4 \right) \left( 4 - 3i \right)}{\left( 4 + 3i \right) \left( 4 - 3i \right) } \right] \right\} \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{104i - 78i^2 }{25} \right) u + \left( \frac{12i - 3i^2 - 16 + 12i}{25} \right) \right] \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{78 + 104i}{25} \right) u + \left( \frac{-13 + 24i}{25} \right) \right] \\ &= \left( 4 + 3i \right) \left[ u^2 + \left( \frac{78 + 104i}{25} \right) u + \left( \frac{39 + 52i}{25} \right) ^2 - \left( \frac{39 + 52i}{25} \right) ^2 + \left( \frac{-13 + 24i}{25} \right) \right] \end{align*}$

$\displaystyle \begin{align*} &= \left( 4 + 3i \right) \left[ \left( u + \frac{39 + 52i}{25} \right) ^2 - \left( \frac{ -1183 + 4056i }{625} \right) + \left( \frac{-325 + 600i}{625} \right) \right] \\ &= \left( 4 + 3i \right) \left[ \left( u + \frac{39 + 52i}{25} \right) ^2 - \left( \frac{-858 + 3456i}{625} \right) \right] \\ &= \left( 4 + 3i \right) \left[ \left( u + \frac{ 39 + 52i }{ 25 } \right) ^2 - \left( \frac{\sqrt{ -858 + 3456i }}{25} \right) ^2 \right] \\ &= \left( 4 + 3i \right) \left( u + \frac{39 + 52i - \sqrt{ -858 + 3456i } }{25} \right) \left( u + \frac{ 39 + 52i + \sqrt{ -858 + 3456i } }{ 25 } \right) \end{align*}$

If you want you CAN get a proper complex number for this square root, but it will be in terms of sines and cosines.

But wolfram alpha gives out this...
$\framebox{3iu^2+26iu+4u^2+3i-4 = (u+(3+4i))((4+3i)u+i)}$
Wolfram doesn't give out a solution for this, and I wanted to know how was it done...
Any idea how it was done?
factorise3iu^2+26iu+4u^2+3i-4 - Wolfram|Alpha

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