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- Newsgroups: comp.sys.hp48
- Path: sparky!uunet!munnari.oz.au!bruce.cs.monash.edu.au!monu6!yoyo.cc.monash.edu.au!quetzal
- From: quetzal@yoyo.cc.monash.edu.au (Quetzal)
- Subject: Re: vector root finder, etc
- Message-ID: <1993Jan13.004254.14911@monu6.cc.monash.edu.au>
- Sender: news@monu6.cc.monash.edu.au (Usenet system)
- Organization: Monash University, Melb., Australia.
- References: <C0qvCv.Hr@news.iastate.edu>
- Date: Wed, 13 Jan 1993 00:42:54 GMT
- Lines: 16
-
- In <C0qvCv.Hr@news.iastate.edu> twh18@isuvax.iastate.edu (Phoenix) writes:
-
- > Hi, I was looking a standard trig book yesterday, and some a
- >problem like "take the cube root of (-1, SQRT 3). I tried to type
- >it in, then discoverd that I cannot put the SQRT in there no matter
- >which mode I use(vector or complex, () or []). The only way I can
- >get a put the SQRT sign in the vector is enter it as (-1, 3), then
- >edit it. Does anyone know any way to enter SQRT 3 directly in there?
-
- Since there is a symbolic in the complex number, you'll need to enclose
- the whole thing in 'these' to get it through the parser. I don't know
- if it works for SQRT signs, my calc isn't with me, but it that doesn't
- work (you *need* the comma between the two components), you'll have to
- enter it as '-1+SQRT 3*i' then do ->NUM on it.
-
- Tim Pickett
-