Thread: complex numbers query

1. complex numbers query

heey so suppose I have 5-5i = 5 root 2 (cosx+isinx)

now i have to equate the roots to eventually get x
is it always the case that I can choose to equate the real or imaginary and it doesn't matter which one? do I always get the same answer??

2. Re: complex numbers query

Originally Posted by sssona09
heey so suppose I have 5-5i = 5 root 2 (cosx+isinx)

now i have to equate the roots to eventually get x
is it always the case that I can choose to equate the real or imaginary and it doesn't matter which one? do I always get the same answer??
$\noindent You don't have to equate real and/or imaginary parts. We know that x is the argument of 5-5i, and we know that the argument of 5-5i is -\frac{\pi}{4} and so x = -\frac{\pi}{4}.$

3. Re: complex numbers query

Originally Posted by sssona09
heey so suppose I have 5-5i = 5 root 2 (cosx+isinx)

now i have to equate the roots to eventually get x
is it always the case that I can choose to equate the real or imaginary and it doesn't matter which one? do I always get the same answer??
$\noindent When you equate real and imaginary parts, you get \cos{x} = \frac{1}{\sqrt{2}} \ (\star) and \sin{x} = -\frac{1}{\sqrt{2}} \ (\star \star). From (\star), x = \frac{\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. And from (\star \star), x = -\frac{3\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. \\\\ You may be confused with this set of solutions but since both (\star) and (\star \star) must hold, then you take the common solutions, ie. x = -\frac{\pi}{4}.$

4. Re: complex numbers query

Originally Posted by 1729
$\noindent You don't have to equate real and/or imaginary parts. We know that x is the argument of 5-5i, and we know that the argument of 5-5i is -\frac{\pi}{4} and so x = -\frac{\pi}{4}.$
how do we know that? by plotting?

5. Re: complex numbers query

for complex number z=a+ib, arg(z) = arctan(b/a) [-pi < arg(z) ≤ pi]
and its clear that 5-5i is in the 4th quadrant, so the argument of 5-5i is arctan(-1) = -pi/4

6. Re: complex numbers query

Originally Posted by 1729
$\noindent When you equate real and imaginary parts, you get \cos{x} = \frac{1}{\sqrt{2}} \ (\star) and \sin{x} = -\frac{1}{\sqrt{2}} \ (\star \star). From (\star), x = \frac{\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. And from (\star \star), x = -\frac{3\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. \\\\ You may be confused with this set of solutions but since both (\star) and (\star \star) must hold, then you take the common solutions, ie. x = -\frac{\pi}{4}.$
thank you so much for helping it makes sense now

7. Re: complex numbers query

Originally Posted by jathu123
for complex number z=a+ib, arg(z) = arctan(b/a) [-pi < arg(z) ≤ pi]
so the argument of 5-5i is arctan(-1) = -pi/4
ohhhh thanksss

8. Re: complex numbers query

Originally Posted by 1729
$\noindent When you equate real and imaginary parts, you get \cos{x} = \frac{1}{\sqrt{2}} \ (\star) and \sin{x} = -\frac{1}{\sqrt{2}} \ (\star \star). From (\star), x = \frac{\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. And from (\star \star), x = -\frac{3\pi}{4} or -\frac{\pi}{4} for -\pi < x \leq \pi. \\\\ You may be confused with this set of solutions but since both (\star) and (\star \star) must hold, then you take the common solutions, ie. x = -\frac{\pi}{4}.$
Hey what if the domain is x between 0 and pi/2
would i say that sinx has no solutions?? both should hold true so..

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