complex numbers query (1 Viewer)

sssona09 · Oct 29, 2017

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??

1729 · Oct 29, 2017

sssona09 said:
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 $x = -\frac{\pi}{4}.$

1729 · Oct 29, 2017

sssona09 said:
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. $x = -\frac{\pi}{4}$.$

sssona09 · Oct 29, 2017

1729 said:
$$\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?

jathu123 · Oct 29, 2017

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

sssona09 · Oct 29, 2017

1729 said:
$$\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

sssona09 · Oct 29, 2017

jathu123 said:
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

sssona09 · Oct 29, 2017

1729 said:
$$\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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complex numbers query (1 Viewer)

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