Re: equation question

• To: mathgroup at smc.vnet.net
• Subject: [mg70079] Re: equation question
• From: ab_def at prontomail.com
• Date: Mon, 2 Oct 2006 00:34:05 -0400 (EDT)
• References: <efdk51\$qg\$1@smc.vnet.net>

dimmechan at yahoo.com wrote:
> Hello.
>
> Consider the following simple examples of FindRoot application.
>
> FindRoot[Sin[x] == 2, {x, I}]
> {x -> 1.5707963267948966 + 1.3169578969248168*I}
>
> FindRoot[Sin[x^2] == 2, {x, I + 1}]
> {x -> 1.3454777060580754 + 0.4894016047219337*I}
>
> FindRoot[Sin[x^2] == 2, {x, 3*I + 2}]
> {x -> 0.3004695589886017 + 2.1914997002654357*I}
>
> Is it possible for FindRoot (or in general in another way) to search
> for solutions
> in the complex plane in an particular domain e.g. searching in the
> domain that
> is made by the lines Re[x]=a1, Re[x]=a2 and Im[x]=b1, Im[b]=b2 ?
>
> I really appreciate any assistance.
>
> Regards
> Dimitris

You can specify a rectangular search region in the FindRoot iterator
(the bottom left and top right corners):

In[1]:= FindRoot[Sin[x^2] - 2, {x, 2 + 3*I, 0, 3 + 4*I}]

Out[1]= {x -> 0.30046956 + 2.1914997*I}

Maxim Rytin
m.r at inbox.ru
***********************************

• Prev by Date: Re: subsets of a set
• Next by Date: Re: DownValues question
• Previous by thread: Re: Initialization cell and button
• Next by thread: Re: DownValues question