# Thread:

1. ## complex no

if z1=10+6i and z2=4+6i if z is any complex no such that the argument of (z-z1)/(z-z2)
is pi/4 then prove that mod(z-7-9i)=3sqrt(2)

can yu solve it graphically

2. Originally Posted by prasum
if z1=10+6i and z2=4+6i if z is any complex no such that the argument of (z-z1)/(z-z2)
is pi/4 then prove that mod(z-7-9i)=3sqrt(2)

can you solve it graphically
Graphically, the argument of the quotient $(z-z_1)/(z-z_2)$ is the angle between the lines $z\to z_1$ and $z\to z_2$. By the theorem about angles in the same segment (or rather its converse), that says that z is on a circle through $z_1$ and $z_2$. Another theorem (the one about the angle at the centre C being twice the angle at the circumference) says that the lines $z_1\to C$ and $z_2\to C$ must be at right angles. It's easy to deduce from that (see the diagram) that C must be at the point 7+9i and that the radius must be $3\sqrt2$. So z lies on the circle $|z-(7+9i)| = 3\sqrt2.$
Attached Thumbnails

#### Search Tags

Copyright © 2005-2013 Math Help Forum. All rights reserved.
{"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": 9, "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": false}, "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": 0.9185511469841003, "perplexity_flag": "middle"}