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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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