IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => University => Topic started by: Syeda Zuriat Zahra on November 21, 2009, 09:47:00 am
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Can anyone solve this question
Find the value of z for which sinz=2i and cosz=2.
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Gimme am hour to get to c pc
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cosz=2
cos(x+iy)=cosxcoshy-isinxsinhy=2 so cosxcoshy=2 (1) and -sinxsinhy=0 and x=n pi but from (1) cosh is positive so cos is positve and x=2npi and from (2) y=cosh^-1(-2)
sinz=2i so sinxcoshy+icosxsinhy=2i so sinxcoshy=0 cosxsinhy=2 (2)
from 1st equationx=npi and from second sinhy=2/cos(n pi) so y =sinh^-1(2/cos(n pi))
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No you are doing it all wrong. i know it a little bit but i don't now how to solve it till the end. I'll give you a hint
cosz=2
cos(x+iy)=2
cosxcoshy-isinxsinhy=2+0i
cosxcoshy=2 -sinxsinhy=0 now i cannot solve further.
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I am so stupid. How did i make that mistake. Will do it when i get up.
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How did you get the value of x? mostly do it by putting sinhy not=0 and sinx=0, x=sin-10
x=npi but this is wrong. how can we put sinhy not=0 because sinh0=0.
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this is weird analysis.
sounds like a different language.
i dont know anything about this
but i just used this on a search engine called Wolfram alpha
lol it solved the question but the input is wrong :(
try it out urself :D
http://www.wolframalpha.com/input/?i=sinz%3D2i+cosz%3D2
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sin npi=0 Try it
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is it something taught to students in maths while obtaining a bachelors degree?
or is it taught to students in maths when they r obtaining a masters degree?
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Actually that is part of further maths at a level probably.
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nope. its at the master level