Precalculus
Topics

# How do you simplify (-6-8i)/(4+12i) ?

Feb 12, 2016

$\frac{- 3 + i}{4}$

#### Explanation:

We will multiply the numerator and the denominator by something called a "complex conjugate." It is the opposite of the current denominator.

So, $4 - 12 i$ is our conjugate.

This leads to $\frac{- 6 - 8 i}{4 + 12 i}$* $\frac{4 - 12 i}{4 - 12 i}$

Foil the numerators to get $- 24 + 72 i - 32 i + 96 {i}^{2}$

Foil the denominators to get $16 - 48 i + 48 i - 144 {i}^{2}$

Remember ${i}^{2} = - 1$

Now, the numerator becomes $- 24 + 40 i + 96 \left(- 1\right)$
---> $- 24 - 96 + 40 i$
---> $- 120 + 40 i$

The denominator becomes $16 - 144 \left(- 1\right)$
--->$16 + 144$
---> $160$

No we have $\frac{- 120 + 40 i}{160}$

Reducing we get $\frac{- 3 + i}{4}$

##### Impact of this question
154 views around the world
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9963648319244385, "perplexity": 3207.2346407726677}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998913.66/warc/CC-MAIN-20190619043625-20190619065625-00030.warc.gz"}