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hello everyone in this video we’re going
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to be evaluating a very interesting very
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complex expression we have I to the
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power I and I is defined as the number
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whose square equals negative one so I is
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the imaginary unit we’re going to talk
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about some interesting formulas uh the
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polar form of a complex number Euler’s
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formula and then I’m also going to be
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showing you a graph all right let’s get
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started so first of all we can use I to
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write complex numbers so X Plus y i If X
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and Y are real numbers Z equals X Plus y
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i is defined as a complex number
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obviously real numbers are also complex
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because if Y is equal to 0 you get X
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which is real so they’re kind of subset
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of complex numbers anyways so the the
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one of the biggest questions that is
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this going to be a complex number as
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well because I is complex a complex
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number to the power complex number is
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that going to be complex or real
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all right let’s go ahead and find out
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first of all let’s talk about Euler’s
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formula so Euler obviously when you look
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up Euler’s formula a couple different
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things are going to come up but one of
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which is very important
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e to the power IX equals cosine X plus I
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times sine X
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now X is measured in radians that
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basically describes the angle from the
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the positive angle you know we’re going
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to talk about it on the unit circle or
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in the coordinate system
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so e to the power IX is an exponential
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but I is imaginary X is real it’s an
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angle measured in radians so on and so
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forth so this allows us to write a
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complex number that’s written in polar
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form
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we can turn it into an exponential form
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and where does this come from obviously
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maybe I shouldn’t use X Plus y i because
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that’s kind of misleading how about we
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use a plus bi so here you can kind of
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see the association between these two
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things one of the things that we’re
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going to Define is going to be the how
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to express a complex number in polar
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form so here’s what we do we’re going to
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define the Theta as the angle measured
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in radians is a positive angle
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it’s counterclockwise and a and b are
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the coordinates so basically a complex
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number can be expressed as a point on
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the coordinate plane or a vector but how
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do you turn this into a polar form you
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can do that by using the following fact
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r as defined as the modulus which is the
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square root of a squared plus b squared
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so if you take out a r then you get a
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over R plus b over r i a over R becomes
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the cosine of an angle and this becomes
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the sign of an angle so on and so forth
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and you can Define the tangent as well
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but let’s go ahead and do this for I now
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I is kind of easy because we can write
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it as 0 plus 1 times I and its modulus
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is 1 because 0 comma one the length is
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going to be one right so here’s how we
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express it here on the coordinate system
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and from here you can actually easily
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find the argument or the angle measured
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in radians in this case it is going to
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be pi over 2. so this is our imaginary
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number this is the imaginary axis this
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is the real axis so if you have any
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complex number whose imaginary part is 0
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then it’s going to be on the real number
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line okay so this is I expressed in
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polar form how do we raise it to the
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power I but we have to turn it into an
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exponential form so let’s go ahead and
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do that so I can be written as since the
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angle is pi over 2 we can write it as
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cosine of pi over 2 plus I times sine pi
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over 2. now in this case I kind of
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ignored the r because R is 1 so we don’t
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have to worry about it so we’re good now
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we’re going to raise this to the power I
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but using Euler’s formula we can write
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this as e to the power I times pi over
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that’s beautiful now we’re going to
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go ahead and raise I to the power I
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which means e to the power I pi over 2
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to the power I
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now
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the exponentiation rules or exponents
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whatever tells us that we’re supposed to
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multiply these two things so it’s going
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to become e to the power I squared pi
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over 2 but we know by definition that I
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squared is equal to negative one so this
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is going to be e to the power negative
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pi over 2. great so that’s the answer
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and as you know this is a real real
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number and you’re like are you for reals
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yes this is a real number interesting
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right and we’re going to look at the
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value of this when I show you the graph
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because we we’re going to talk about
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some solutions right so real and complex
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Solutions but we could also do the
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following I cannot be written in a
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single way you could also write since pi
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over 2 can also be expressed as pi over
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2 plus
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at even multiple of Pi which is
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um or integer multiple of 2 pi so we can
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also write I this way
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if you think about it we can basically
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add multiples of 2 pi to it because it’s
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not going to matter it’s just going to
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give us four more rotations and this can
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be written as e to the power I times pi
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over 2 plus 2 pi n and then when you
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raise this is I by the way when you
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raise it to the power I it’s just going
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to become e to the power I squared times
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this and then you’re just going to place
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I squared with negative 1 and this is
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going to give you e to the power
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negative pi over 2 minus 2 pi n but
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negative n is just another integer so we
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can call that K and this can be written
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as e to the power negative pi over 2
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plus 2 pi K so this is just another way
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to write it in more general form I to
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the power I and if you replace K with 0
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you get the first solution if you
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replace K with one so on and so forth
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obviously there are infinitely many
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solutions as long as K is an integer
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okay great now here’s one thing that I
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want to talk about but first I want to
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give you a little bit of something so
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we’re basically looking at the following
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right we have y equals x to the power x
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and we found that I to the power I is
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equal to e to the power negative pi over
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so let’s just go ahead and stick with
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this one for now okay so this indicates
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that hey if this is equal to e to the
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power negative pi over 2 we’re supposed
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to get x equals I from here right
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because I satisfied this equation at
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least that’s one of the solutions so I
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to the power I is equal to this number
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right here and that’s a real number but
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I is not a real number so what is the
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issue here the issue is the derivative
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the maxima and minimum and so on and so
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forth so let’s go ahead and align both
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sides of this function this becomes Ln x
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to the X and then Ln y becomes X Ln X
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and I’m going to go ahead and
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differentiate by using chain rule y
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Prime over Y is derivative of x times Ln
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X plus the derivative of Ln x times x
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this becomes 1 and y Prime becomes y
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times this which is X to the x times Ln
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X plus 1. when you set it equal to zero
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you get Ln x equals negative 1 which
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means x equals one over e and then if
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you plug in 1 over e you get 1 over e
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comma 1 over e to the power one over e
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which is e to the power negative one
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over e so that point is actually a
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minimum point on the graph because if
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you remember the graph of y equals x to
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the power x it kind of looks like this
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and it makes a minimum at this point but
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what happens when you have a horizontal
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line that’s what we’re going to look at
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right now let’s go ahead and take a look
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and here’s our graph of y equals x to
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the power x and the solution
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for I to the power I so this is equal to
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I to the I so we’re expecting to get an
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x equals I here but there is no
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intersection point because I is not a
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real number and obviously this value
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here is less than the minimum y value
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that you see here and this brings us to
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the end of this video thanks for
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watching I hope you enjoyed it please
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let me know don’t forget to comment like
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And subscribe I’ll see you tomorrow with
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another video Until then be safe take
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care and bye bye