Register to reply

Complex equation

by powers
Tags: complex, equation
Share this thread:
powers
#1
Jan2-13, 01:53 PM
P: 2
Hi, I need a little help

I need to find solution for this equations:

[itex]\frac{Z-a}{Z-b}[/itex]=K[itex]e^{ąjθ}[/itex]

The Z is unknown and it is the complex number. The a and b is known and they are also complex numbers. K is the real number.

I know that for [itex]-90^{°}[/itex]<θ<[itex]90^{°}[/itex] the graph in the complex plane is circle, for [itex]-45^{°}[/itex]<θ<[itex]45^{°}[/itex] the graph in the complex plane is in shape of "tomato" and for [itex]-135^{°}[/itex]<θ<[itex]135^{°}[/itex] is shape of "lens", but I don't know how to solve it.

Sorry if my post is in wrong area.

Thanks for help.
Phys.Org News Partner Mathematics news on Phys.org
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
jfgobin
#2
Jan2-13, 03:04 PM
P: 90
Leaving it to you the conditions of existence:

[itex]Z=\frac{a-b.K.\textrm{e}^{ \pm j \theta }}{1-K.\textrm{e}^{ \pm j \theta }}[/itex]
powers
#3
Jan2-13, 04:06 PM
P: 2
In that way I got only the one solution, where are the other?
For example, let's put b=0, K=1, theta=45°, with above formula we got only the one solution, but there is more than one solution...

Char. Limit
#4
Jan12-13, 04:02 PM
PF Gold
Char. Limit's Avatar
P: 1,951
Complex equation

How do you only get one solution when there's clearly a [itex]\pm[/itex] in his answer?


Register to reply

Related Discussions
Complex equation Calculus & Beyond Homework 8
Heat equation solving quadratic equation with complex numbers Calculus & Beyond Homework 1
Complex equation Calculus 9
How to fit a complex equation? General Math 3
Complex cosine equation (complex analysis) Calculus & Beyond Homework 2