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Hey guys,
Having finished matrices I've moved on to complex numbers. It's been going well but I got stuck on one question
Show that there is only one value of a for which
,And find this value.
I believe that
However, I get the argument of this to be:
Which is equal to:
But if I equate that to
I can only get:
Which, of course, has two solutions.
Am I missing something?
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Hi Au101,
For a = 1,
z = 1+i ; arg(z) = pi/4.
For a = -6
z = -0.75 - 0.75i ; arg(z) = pi + pi/4
Hence, if you need the phase to be in between -pi and pi, a=1 only
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Hmmm, I see, thanks a lot gAr. Although, I still don't know quite how I would set out a proof that there can be only one value of a, especially - as the question implies - without finding a first. Could you, or anybody else, advise?
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You're welcome!
Show that there is only one value of a for which arg(z)=pi/4
I showed the same. For arg(z) = pi/4, a=1.
The question is not asking you to prove that the quadratic equation has only one solution!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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Ohhh okay, fair enough .
Thanks a lot!
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You're welcome!
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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