Math Contest #21 [3 SBI]
Here you can keep your brain fit by solving math related problems and also earn SBI or sometimes other rewards by doing so.
The problems usually contain a mathematical equation that in my opinion is fun to solve or has an interesting solution.
I will also only choose problems that can be solved without additional tools(at least not if you can calculate basic stuff in your head), so don't grab your calculator, you won't need it.
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Rules
No upvote, No resteem, No follow required!
I will give the SBI(s) randomly to any participants.
You have 4 days to solve it.
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Problem
Today it's time for a normal equation again:
Solve for x!
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To everyone who already participated in a past contest, come back today and try a new problem(tell me if you don't want to be tagged):
@addax @ajayyy @athunderstruck @bwar @contrabourdon @crokkon @fullcoverbetting @golddeck @heraclio @hokkaido @iampolite @kaeserotor @masoom @mmunited @mobi72 @mytechtrail @ninahaskin @onecent @rxhector @sidekickmatt @sparkesy43 @syalla @tonimontana @vote-transfer @zuerich
In case no one gets a result(which I doubt), I will give away the prize to anyone who comments.
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@contrabourdon sponsors my contests with 2 STEEM weekly.
You can support him by using a witness vote on untersatz, so he can further support this and other contests.
E^x=(1+i)
Ln(e^x)=ln(1+i)
x=ln(1+i)
====
This is as far as I got by myself
Google says ln(1+i) = 0.34657359+0.785398163i
Actually I want you try to find the log of complex numbers yourself.
It isn't that hard when you keep euler's formula in mind:
e^x+iy = e^x *(cos(y)+i*sin(y))
Write 1+i=z in exponential form:
|z|=sqrt(1^2+1^2)=sqrt(2)
tan(phi)=1/1, phi=arctan(1)+2 * k * Pi=0.7854+2 * k * Pi
z=1+i=sqrt(2) * e^(i * [0.7854+2 * k * Pi]
x = ln(1+i) = ln{ sqrt(2) * e^(i * [0.7854+2 * k * Pi] } = ln(sqrt(2)) + i * [0.7854+2* k * Pi]
x = 0.3466 + i(0.7854+ 2 * k * Pi)
With k in N
Sorry for formatting btw, but the * sign makes everything italic^^
You can use \* and markdown won't make it italic.