# Thread:

1. ## Complex Number Proof

For the complex equation $z^4 = cosx + isinx$, prove that the sum of the four solutions is always zero, no matter what size $x$ is.

This is what I've done so far.

$Z = (cis(x + 360k))^1^/^4<br /> = cis(x/4 + 90k)$

Can someone please teach me how to do the rest? Thanks in advance

2. Rewrite it as $\displaystyle z^4 = e^{ix}$.

This means that the first fourth root is $\displaystyle z = \left(e^{ix}\right)^{\frac{1}{4}} = e^{i\frac{x}{4}} = \cos{\left(\frac{x}{4}\right)} + i\sin{\left(\frac{x}{4}\right)}$.

The other fourth roots are evenly spaced around a circle, so differ by an angle of $\displaystyle \frac{\pi}{4}$. What are the other solutions? What do you get when you add them together?

3. A little twist on ProveIt's approach:

Let $w = \cos x + i \sin x$, so the equation can be written
$z^4 - w = 0$.

If the roots are $r_1, r_2, r_3, r_4$, then we must have
$z^4 - w = (z - r_1) (z - r_2) (z - r_3) (z - r_4)$.

If we expand the right-hand side of this equation, what can we say about the coefficient of $z^3$?

« graph | example »
{"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": 11, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.920246422290802, "perplexity_flag": "head"}