# Thread:

1. ## factorising over complex numbers:

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

Thanks

2. ## Re: factorising over complex numbers:

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

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

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

3. ## Re: factorising over complex numbers:

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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8751527667045593, "perplexity_flag": "middle"}