Elementary Functions - Complex Analysis

Again we can use Euler's formula or the complex unit circle.

The trick here is to realize that ,where is real of course, cannot be negative.

thus we have the following. . Thus we have the two

equations below.

now the top equation above has a real part and an imaginary part. Since is real we set it equal to the real part. we now have the two equations below.

These two equations can be solved with pre-calculus and trigonometry skills.

now if your familiar with then you know it is a periodic function.

is not the only solution to

any angle co-terminal with is also a solution. Thus for the final answer we have...

we could have solved it a lot quicker if we used the complex unit circle to solve

If you know how to use the trigonometric unit circle to find a trig value then

you already in a way know how to find the answer to any . The only

difference with the complex unit circle is that they component is imaginary.

So to find the answer of you just move counter clockwise from the

positive axis by and take that value. which in this case is .

It is worth noting that every point on the complex unit circle has two

components, a real and imaginary part. At the (real part) is zero, and

the (imaginary) is .

the or imaginary axis is the only place where is pure imaginary.

The angles that put you there are

,

and those angles that are co-terminal with them.

Thus....

Now the picture i posted with this problem was the best i could find.

It is a little misleading.

The the picture refers to are only those on the complex unit circle.

The answer to the question needs to be any point on the complex

plane. we want the entire (complex) axis in our solution.

above my is not the same as in the picture. The part in the line just

above causes the complex point to move up or down the (complex) axis to

infinity etc. thus the whole axis.

For and its co-terminal angles and in the interval

the positive imaginary axis is covered. For and its co-terminal

angles and in the interval the negative imaginary axis is

covered. Thus the final solution is below.

First replace the in the above equation with .

The is included since the circle is periodic and multiples of are

co-terminal with .

Now take the fourth root of both sides of the equation.

now simplify...

by inspection of the equation above we can see that the solutions are given by

multiples of . In other words we take the complex unit circle and start

at the axis and move counter clockwise by angle increments of .

This forms the picture below.

As seen above the picture forms a square with the starting vertex at .

This shape is not an accident. For example the cube roots of unity form an

equilateral triangle with starting vertex at .

Below are the solutions to the cube roots of unity. One can recognize those

values from the triangle.

This technique is a quick way to find many of the lower number roots of unity.

It is a good approach for the visual learner. Any nth root of unity will form a

regular n-gon inscribed in the unit circle with the starting vertex at .

For more complex root equations, one would start in the same way but use

Euler's equation to finish the problem.

Finally, the reason I did not choose

as the correct answer is that it produces infinite and duplicate answers which

can be boiled down to just

first we get the right hand side of the equation into exponential form.

thus we now have...

and

the first half of the above equations implies .

the second equation implies but we must be careful.

should make you think of the complex unit circle, and it is periodic.

we can wheel around by from and be co-terminal with

infinitely many times. Thus

so the final solution is...

first we have to take the above equation and split it into its real and imaginary

parts.

Now we take the partial derivatives of and .

If you are not already familiar here are the CR equations below.

and

by substitution we get the following system of equations.

now lets simplify....

can never equal zero so we can divide by it on both sides of both equations

without accidentally dividing by zero.

Further can be any real and this division still holds. The point being in our

solution can be any real.

sometimes sine and cosine can be zero so we cannot divide by them on both

sides. We can collect the like terms and set the equations equal to zero.

and like the the twos divide away.

Now we have established can be any real. Now we just need a or more

than one or infinitely many values that can solve that system.

Unfortunately there does not exist any that can simultaneously solve that

system.

Thus for every on the complex plane the CR equations are not satisfied.

Every point on the complex plane has an component and a component in

the form or in other words . That cannot satisfy the

CR equations if the part does not satisfy the CR equations.

Thus the function does not satisfy the CR equations on any point in the

complex plane.

The first task is to split the function into its real and imaginary parts.

The step above is algebra.

The step above is using Euler's formula.

now that we have and we take the partial derivatives. The product rule is used to take the derivatives.

then with algebra the above partial derivative of U will simplify. The rest of the partials are done in similar manner i will list the rest in their final simplified form.

The CR equations are below.

and

By inspection of the above partials, one can see that the equations are identical.

Thus like when you have:

the equations are true for all and .

Thus the CR equations are satisfied for the whole complex plane.

Once again this can be finished with either Euler's formula or the complex unit circle.

should by now in this practice test remind you of the complex unit circle. The radius is the magnitude of every point on that circle. So for any or in this case any ...

thus....

above are the steps to get the magnitude of the left side of the inequality.

above are the steps to get the magnitude of the right side of the inequality.

Thus becomes...

now we do algebra...

with this last inequality you can graph the right hand side and see that it is

always or greater, or you can reason it this way...

Thus the inequality is true for all complex numbers.

where is the complex number such that

Converting into polar coordinates

This gives us

The first solution method uses Euler's formula.

The second solution uses the complex unit circle. It begins in the same manner.

Just as was done in Trigonometry, you swing an angle from the positive axis. In this case is , and the radius is one. An angle of degrees starting from the positive axis will land you at on the negative axis. The value for . It is worth mentioning that the complex unit circle has at each point two components. At the angle equal , the component is and the component is . Thus...

Also, the notation can be a little confusing it seems as if at one point and at another point . and also at one point , and at another point . This is because there are two .

and

Again we can use Euler's formula or the complex unit circle.

The trick here is to realize that ,where is real of course, cannot be negative.

thus we have the following. . Thus we have the two

equations below.

now the top equation above has a real part and an imaginary part. Since is real we set it equal to the real part. we now have the two equations below.

These two equations can be solved with pre-calculus and trigonometry skills.

now if your familiar with then you know it is a periodic function.

is not the only solution to

any angle co-terminal with is also a solution. Thus for the final answer we have...

we could have solved it a lot quicker if we used the complex unit circle to solve

If you know how to use the trigonometric unit circle to find a trig value then

you already in a way know how to find the answer to any . The only

difference with the complex unit circle is that they component is imaginary.

So to find the answer of you just move counter clockwise from the

positive axis by and take that value. which in this case is .

It is worth noting that every point on the complex unit circle has two

components, a real and imaginary part. At the (real part) is zero, and

the (imaginary) is .

the or imaginary axis is the only place where is pure imaginary.

The angles that put you there are

,

and those angles that are co-terminal with them.

Thus....

Now the picture i posted with this problem was the best i could find.

It is a little misleading.

The the picture refers to are only those on the complex unit circle.

The answer to the question needs to be any point on the complex

plane. we want the entire (complex) axis in our solution.

above my is not the same as in the picture. The part in the line just

above causes the complex point to move up or down the (complex) axis to

infinity etc. thus the whole axis.

For and its co-terminal angles and in the interval

the positive imaginary axis is covered. For and its co-terminal

angles and in the interval the negative imaginary axis is

covered. Thus the final solution is below.

First replace the in the above equation with .

The is included since the circle is periodic and multiples of are

co-terminal with .

Now take the fourth root of both sides of the equation.

now simplify...

by inspection of the equation above we can see that the solutions are given by

multiples of . In other words we take the complex unit circle and start

at the axis and move counter clockwise by angle increments of .

This forms the picture below.

As seen above the picture forms a square with the starting vertex at .

This shape is not an accident. For example the cube roots of unity form an

equilateral triangle with starting vertex at .

Below are the solutions to the cube roots of unity. One can recognize those

values from the triangle.

This technique is a quick way to find many of the lower number roots of unity.

It is a good approach for the visual learner. Any nth root of unity will form a

regular n-gon inscribed in the unit circle with the starting vertex at .

For more complex root equations, one would start in the same way but use

Euler's equation to finish the problem.

Finally, the reason I did not choose

as the correct answer is that it produces infinite and duplicate answers which

can be boiled down to just

first we get the right hand side of the equation into exponential form.

thus we now have...

and

the first half of the above equations implies .

the second equation implies but we must be careful.

should make you think of the complex unit circle, and it is periodic.

we can wheel around by from and be co-terminal with

infinitely many times. Thus

so the final solution is...

first we have to take the above equation and split it into its real and imaginary

parts.

Now we take the partial derivatives of and .

If you are not already familiar here are the CR equations below.

and

by substitution we get the following system of equations.

now lets simplify....

can never equal zero so we can divide by it on both sides of both equations

without accidentally dividing by zero.

Further can be any real and this division still holds. The point being in our

solution can be any real.

sometimes sine and cosine can be zero so we cannot divide by them on both

sides. We can collect the like terms and set the equations equal to zero.

and like the the twos divide away.

Now we have established can be any real. Now we just need a or more

than one or infinitely many values that can solve that system.

Unfortunately there does not exist any that can simultaneously solve that

system.

Thus for every on the complex plane the CR equations are not satisfied.

Every point on the complex plane has an component and a component in

the form or in other words . That cannot satisfy the

CR equations if the part does not satisfy the CR equations.

Thus the function does not satisfy the CR equations on any point in the

complex plane.