# 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

#### 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

03.11.2019 00:27
0

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)+isin(y))

03.11.2019 09:28
0

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

03.11.2019 12:23
1

Sorry for formatting btw, but the * sign makes everything italic^^

03.11.2019 12:25
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You can use \* and markdown won't make it italic.

03.11.2019 13:05
0