We're asked to use Euler's Formula
to find the two roots or two complex solutions
of Z squared equals two i.
Euler's Formula expresses a complex number
in the form of X plus yi in exponential form
which is the form r times e raised to the power of i theta
where r is the modulus, or distance to point
x comma y is from the origin.
This is the same r in polar form,
and the angle theta is the angle in standard form
or standard position whose terminal side contains
the point x comma y.
Because we're looking for the two roots
or two complex solutions, we need to find
two Euler forms in exponential forms
of the complex number two i using
two consecutive coterminal angles.
And for this example we will use
the least positive coterminal angles.
In general, to find n roots we provide
in Euler forms or exponential forms
of the number using n consecutive coterminal angles.
So for the next step we want to plot two i
on the complex coordinate plane,
and since two i is equal to zero plus two i,
it's graphed as the point zero comma two
shown here in blue.
Because the point is on the vertical axis
we can quickly see it is two units from the origin
which is why r is equal to two.
But, of course, we can always use
the formula r equals the square root
of the quantity x squared plus y squared
which also gives us r equals two.
And now let's find the least positive angle
where the initial side is
along the positive horizontal axis,
and the terminal side passes through the point
zero comma two which we can quickly see is
equal to 90 degrees or pi over two radians
or 1/2 pi radians which we'll use
for our first angle theta.
And then to find the next positive coterminal angle,
we add two pi radians to 1/2 pi radians
which gives us 5/2 pi radians.
So now we have the information we need
to write two exponential forms for two i
by substituting r equals two and theta equals
1/2 pi radians as well as theta equals 5/2 pi radians
in the exponential form r times e raised
to the power of i theta which is shown here.
The next step to find the two square roots
or two complex solutions to the given equation,
we need to find the square root or 1/2 power
of these two exponential forms or Euler forms for two i.
And let's show this on the next slide.
We will call the two roots or two solutions
z sub one and z sub two where we have raised
both exponential forms for two i to the 1/2 power.
Simplifying, we raise to the 1/2 power
which gives us the square root of two,
and then for e raised to the power of 1/2 pi i,
we multiply the exponent by 1/2 which gives us e raised
to the power of 1/4 pi i.
Because the angle theta is a nice angle meaning
a multiple of 30 degrees or 45 degrees,
we write the exponential form back in polar form
which is this form here where r is equal
to the square root of two and theta's equal
to 1/4 pi radians.
And now to evaluate this by hand,
we need to find the exact trig function values
for a cosine 1/4 pi and sine 1/4 pi which we can do
using the unit circle or reference triangles.
Using the unit circle, remember x equals cosine theta
and y equals sine theta, and here we have
angle pi over four radians or 1/4 pi radians.
Notice both sine and cosine of 1/4 pi radians is
equal to square root two divided by two.
Then we can also find these trig functions values using
the 45-45-90 reference triangle.
And then lastly we should just
distribute square root two.
Remember square root two times square root two is
equal to two, two divided by two is equal to one.
This product simplifies to one plus i
which is the first root or first complex solution.
And then for the second root,
when multiplying the exponent e by 1/2,
we get e raised to the power of 5/4 pi i.
Converting to polar form, we now need to find
the cosine and sine function values for 5/4 pi radians
which we can see here are both equal
to negative square root two divided by two.
So distributing square root two, notice now
we get negative one minus i.
And now we have our two roots or two complex solutions
which are one plus i and negative one minus i.
And notice how for these complex numbers,
both x and y are real, which means
we could've also evaluated this on our graphing calculator.
So let's also show this.
Notice how on the graphing calculator,
we could enter the exponential form
of the two roots to get the two roots
in the form of x plus y i.
I hope you found this helpful.