Complex Analysis - Elementary Functions

$|e^{Z^{2}}|\leq\ e^{|Z|^{2}}$

Is the above inequality true?

Yes it is true for the whole complex plane.

No it is not true. The statement is false for the entire complex plane.

The truth of the statement depends on the value of . In other words it is true for some restricted domain of the complex plane.

It is true when:

The truth of the statement depends on the value of . In other words it is true for some restricted domain of the complex plane.

It is true only for the real line.

The truth of the statement depends on the value of . In other words it is true for some restricted domain of the complex plane.

It is true only for the pure imaginary line.

Explanation

$|e^{Z^{2}}|=|e^{(x+yi)^{2}}|=|e^{(x^{2}-y^{2})+(2xy)i}|$

$|e^{(x^{2}-y^{2})+(2xy)i}|=|e^{x^2-y^2}|*|e^{2xyi}|=e^{x^2-y^2}*1=e^{x^2-y^2}$

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

$e^{|Z|^{2}}=e^{|x+yi|^{2}}=e^{(\sqrt{x^2+y^2})^{2}}=e^{x^2+y^2}$

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

Thus $|e^{Z^{2}}|\leq\ e^{|Z|^{2}}$ becomes...

$e^{x^2-y^2}\leq e^{x^2+y^2}$

now we do algebra...

$e^{x^{2}}/e^{y^{2}}\leq e^{x^{2}}*e^{y^{2}}$

$e^{x^{2}}\leq e^{x^{2}}*e^{y^{2}}*e^{y^{2}}$

$1\leq e^{y^{2}}*e^{y^{2}}$

$1\leq e^{2y^{2}}$

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...

$y^{2}\geqslant0$

$2y^{2}\geqslant0$

$e^{2y^{2}}\geq e^{0}$

$e^{2y^{2}}\geq 1$

Thus the inequality is true for all complex numbers.

Find all values of such that... $e^{Z}=-2$

$ln(2)+(2n+1)\pi i$

$n\in \mathbb{Z}$

$ln(2)+\pi i$

$ln(2)-\pi i$

$ln(2)+(2n)\pi i$

$n\in \mathbb{Z}$

Explanation

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

$e^{Z}=e^{x+yi}=e^{x}*e^{yi}=e^{x}(cos[y]+i sin[y]) = -2$

The trick here is to realize that $e^{x}$ ,where is real of course, cannot be negative.

thus we have the following. $e^{x}*e^{yi}=2*-1$ . Thus we have the two

equations below.

$e^{x}=2$

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.

$e^{x}=2$

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

$y=\pi$

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

$\pi$ is not the only solution to

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

$y=(2n+1)\pi$

$Z=x+yi=ln(2)+(2n+1)\pi i$

$n\in \mathbb{Z}$

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

$e^{yi}=-1$

Eulersfunitc

If $e^{Z}$ is pure imaginary what restriction is placed on ?

Eulersfunitc

hint: where is the only place(s) on the complex unit circle where is pure imaginary and what is the angle(s) that put you there.

$\varphi=\pi/2+n\pi$

$n\in \mathbb{Z}$

$\varphi=\pi/2$

$\varphi=-\pi/2$

$\varphi=\pi/2+n\pi/2$

$n\in \mathbb{Z}$

$y\in \mathbb{R}$

Explanation

Eulersfunitc

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

The angles that put you there are

$\pi/2,-\pi/2$ ,

and those angles that are co-terminal with them.

Thus....

$\varphi=\pi/2+n\pi$

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.

$e^{Z}=e^{x+yi}=e^{x}*e^{yi}=e^{x}*e^{i( (\pi /2)+n\pi) }$

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 $y=\pi/2$ and its co-terminal angles and in the interval $(-\infty ,\infty )$

the positive imaginary axis is covered. For $y=-\pi/2$ and its co-terminal

angles and in the interval $(-\infty ,\infty )$ the negative imaginary axis is

covered. Thus the final solution is below.

$\varphi=\pi/2+n\pi$

$n\in \mathbb{Z}$

What does $e^{Z+\pi i}$ equal?

$e^{-Z}$

$-e^{-Z}$

$e^{x+yi}$

Explanation

$e^{Z+\pi i}=e^{Z}*e^{\pi i}=e^{Z}*-1=-e^{Z}$

$f(x+yi)=e^{x-yi}$

where in the complex plane does the above function satisfy the Cauchy-Riemann equations.

It does not satisfy the CR equations at any point on the complex plane.

It satisfies the CR equations on the real line of the complex plane.

In other words the axis or

It satisfies the CR equations at:

$y=n\pi$

$n\in \mathbb{Z}$

It satisfies the CR equations at:

$y=n\pi/2$

$n\in \mathbb{Z}$

It satisfies the CR equations everywhere on the complex plane.

Explanation

$f(x+yi)=e^{x-yi}$

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

parts.

$e^{x-yi}=e^{x}*e^{-yi}=e^{x}(cos(-y)+i \sin (-y))$

$e^{x}(cos(-y)+i \sin (-y))=e^{x}(\cos (y)-i\sin (y))$

$U(x,y)=e^{x}\cos \left (y \right )$

$V(x,y)=-e^{x}\sin \left ( y\right )$

Now we take the partial derivatives of and .

$U_{}x(x,y)=e^{x}\cos \left (y \right )$

$V_{}y(x,y)=-e^{x}\cos\left ( y\right )$

$U_{}y(x,y)=-e^{x}\sin\left (y \right )$

$V_{}x(x,y)=-e^{x}\sin\left ( y\right )$

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

$U_{x}=V_{y}$ and $U_{y}=-V_{x}$

by substitution we get the following system of equations.

$e^{x}\cos \left (y \right )$ $=-e^{x}\cos\left ( y\right )$

$-e^{x}\sin\left (y \right )$ $=e^{x}\sin\left ( y\right )$

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.

$\cos \left (y \right )$ $=-\cos\left ( y\right )$

$-\sin\left (y \right )$ $=\sin\left ( y\right )$

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.

$2\cos \left (y \right )=0$

$2\sin\left (y \right )=0$

and like the the twos divide away.

$\cos \left (y \right )=0$

$\sin\left (y \right )=0$

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.

Compute $e^{\left(\frac{2+\pi i}{4}\right)}$

$\sqrt{\frac{e}{2}} (1+i)$

$\sqrt{2e}(1+i)$

$\sqrt{e}(1+i)$

$\frac{\sqrt{2}}{2}(1+i)$

None of the choices

Explanation

$e^{\frac{2+\pi i}{4}} = e^{\frac{2}{4}}*e^{\frac{\pi i}{4}} = \sqrt{e} * \frac{\sqrt{2}}{2}(1+i) = \sqrt{\frac{e}{2}}(1+i)$

$e^{Z}=1+\sqrt{3}i$

Solve the above equation for all values of .

$ln(2)+(\pi /3+2\pi n)i$

$n\in \mathbb{Z}$

$ln(2)+(-\pi /3+2\pi n)i$

$n\in \mathbb{Z}$

$ln(2)-(\pi /3+2\pi n)i$

$n\in \mathbb{Z}$

$ln(2)+(\pi /6+2\pi n)i$

$n\in \mathbb{Z}$

$ln(2)+(-\pi /6+2\pi n)i$

$n\in \mathbb{Z}$

Explanation

$e^{Z}=1+\sqrt{3}i$

first we get the right hand side of the equation into exponential form. $re^{\theta i}$

$r=\sqrt{x^2+y^2}=\sqrt{1^2+\sqrt{3}^2}=2$

$\theta=\tan^{-1}(y/x)=\tan^{-1}(\sqrt{3})=\pi/3$

thus we now have...

$e^{Z}=2e^{\pi i/3}$

$e^{Z}=e^{x}*e^{yi}=2e^{\pi i/3}$

$e^{x}=2$ and $e^{yi}=e^{\pi i/3}$

the first half of the above equations implies .

the second equation implies $y=\pi/3$ but we must be careful.

$e^{yi}$ should make you think of the complex unit circle, and it is periodic.

we can wheel around by $2\pi n$ from $\pi/3$ and be co-terminal with $\pi/3$

infinitely many times. Thus $y=(\pi /3+2\pi n)$

so the final solution is...

$ln(2)+(\pi /3+2\pi n)i$

Eulersfunitc

Use the complex unit circle to find the value of $e^{i\pi /2}$

Explanation

Eulersfunitc

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 $e^{\varphi i}$ . The only

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

So to find the answer of $e^{i\pi /2}$ you just move counter clockwise from the

positive axis by $90^{0}$ 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 $90^{0}$ the (real part) is zero, and

the (imaginary) is .

$f(Z)=e^{Z^{2}}$

$f(x+iy)=e^{(x+iy)^{2}}$

Where in the complex plane does the above function satisfy the Cauchy-Riemann equations.

Everywhere. In other words the function satisfies the CR equations on the whole complex plane.

Nowhere. In other words there is not even one point on the complex plane where this function satisfies the CR equations.

the CR equations are satisfied by this function only at:

In other words only on the axis that is the real line.

The CR equations are only satisfied at:

In other words only at the pure imaginary axis.

The CR equations are only satisfied at the origin.

In other words at on the complex plane.

Explanation

$f(x+iy)=e^{(x+iy)^{2}}$

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

$e^{(x^2-y^2)+(2xy)i}=e^{x^2-y^2}*e^{2xyi}$

The step above is algebra.

$e^{x^2-y^2}*e^{2xyi}=e^{x^2-y^2}*(cos(2xy)+i \sin(2xy))$

The step above is using Euler's formula.

$U(x,y)=e^{x^2-y^2}cos(2xy)$

$V(x,y)=e^{x^2-y^2}sin(2xy)$

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

$U_{x}(x,y)=2xe^{x^2-y^2}*cos(2xy)+e^{x^2-y^2}*(-sin(2xy)*2y)$

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.

$U_{x}(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))$