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2d
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Ah okay. And I suppose if $iz=\log(4\pm\sqrt{15})$, the $i$ becomes interchangeable on both sides?
2d
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
The quadratic I got is $e^{2iz}-8e^{iz}+1$. Completing the square gives $e^{iz}=4\pm i\sqrt{15}$ right? The answer lists $\sqrt{15}$ not $i\sqrt{15}$
2d
accepted Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
2d
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
@Integrator huh?
2d
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Thanks. When I complete the square I get $iz=\ln(4\pm i\sqrt{15})$, but the answer listed is $i\ln(4\pm\sqrt{15})$. Is it acceptable to extract out the $i$ like that?
2d
asked Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Nov
21
accepted Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
A few people seem to have substituted $\pi - \theta$ for $\theta$ - which I can kind of understand (it seems kind of arbitrary). I'm not sure how to plug in your substitution, though...
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Shouldn't it be $\pi \operatorname{cosec} \theta \space d\theta$?
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Ah okay. Finally, why does $\int f(a + b-t)dt = - \int f(a + b -t)dt$?
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
How does $\frac{dx}{dt}=1$? And why does $\int f(x)dx=-\int f(a+b-t)dt$? Could you add a few more steps if that's okay? :)
Nov
21
asked Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Nov
19
comment Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
Also - why take $C$ as $0$? Is that an assumption?
Nov
19
comment Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
This is a fantastic answer - I get it now. Thanks!
Nov
19
accepted Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
Nov
19
comment Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
Okay, but where does $y$ go in step 2?
Nov
19
comment Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
Why pick $e^{\int1dx}$ as the multiplier? Does $ye^{\int1dx}=0$?
Nov
19
asked Find the equation of a curve where $\frac{dy}{dx}=2x+y$ at all points
Nov
17
accepted Reduce $\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$ to a homogenous equation by substituting $x=X-1$ and $y=Y+3$
Nov
16
comment Reduce $\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$ to a homogenous equation by substituting $x=X-1$ and $y=Y+3$
@user48481MirkoSwirko Thanks, but I still can't arrive at that answer :/