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Oct 25th, 2006, 04:16 AM
#1
Thread Starter
New Member
confirmation on complex numbers
the question is; solve
z^2 + (2+4i)z -11 -2i = 0
well i did it and my answer is z= 6 + 2i and z= -6 -2i which i know is WRONG right?
BUT say if it is CORRECT, is that my final answer? or do i have to sub that in with something because i am sure you have to.
So can someone give me the correct answer and the steps involve as well as i want to crack this thing once and for all.
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Oct 25th, 2006, 10:52 AM
#2
Re: confirmation on complex numbers
 Originally Posted by vixity
the question is; solve
z^2 + (2+4i)z -11 -2i = 0
well i did it and my answer is z= 6 + 2i and z= -6 -2i which i know is WRONG right?
BUT say if it is CORRECT, is that my final answer? or do i have to sub that in with something because i am sure you have to.
So can someone give me the correct answer and the steps involve as well as i want to crack this thing once and for all.
The correct answer is
z = 2 - i
and
z = -4 -3i
EDIT:
Hang on there while I try to find a useful link...
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
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Oct 25th, 2006, 10:53 AM
#3
Re: confirmation on complex numbers
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
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Oct 26th, 2006, 12:27 AM
#4
Thread Starter
New Member
Re: confirmation on complex numbers
 Originally Posted by krtxmrtz
The correct answer is
z = 2 - i
and
z = -4 -3i
EDIT:
Hang on there while I try to find a useful link...
yeah and what do you do now with the two answers for z. Do you sub it in the question?cause if you do you solve for i correct?
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Oct 26th, 2006, 02:00 AM
#5
Addicted Member
Re: confirmation on complex numbers
The answers for z are the answers to the whole question. i is a constant, you can't solve for it.
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Oct 26th, 2006, 04:39 AM
#6
Re: confirmation on complex numbers
And just to be sure it is clear, i is used to indicate imaginary numbers. It is defined as sqrt(-1). So you can't "solve" for it, it would be like solving for 6.
zaza
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Oct 26th, 2006, 05:24 AM
#7
Re: confirmation on complex numbers
And, as a math teacher I had used to say, if you don't know how to calculate sqr(-1) then what you do is
"you call the problem the solution", so sqr(-1) = i and that's all there is to that.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
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