# Graphs of complex functions

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How do you draw the graph of the modulus of e^z = 1?

This is what I have tried so far:

\[e^z = e^x e^i^y\]

If you take the modulus squared, you get:

\[e^2^x\]

and then square-rooting it gives just \[e^x\] .

So I know that x = 0, but what does y equal?

This is what I have tried so far:

\[e^z = e^x e^i^y\]

If you take the modulus squared, you get:

\[e^2^x\]

and then square-rooting it gives just \[e^x\] .

So I know that x = 0, but what does y equal?

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#2

(Original post by

How do you draw the graph of the modulus of e^z = 1?

This is what I have tried so far:

\[e^z = e^x e^i^y\]

If you take the modulus squared, you get:

\[e^2^x\]

and then square-rooting it gives just \[e^x\] .

So I know that x = 0, but what does y equal?

**IDontKnowReally**)How do you draw the graph of the modulus of e^z = 1?

This is what I have tried so far:

\[e^z = e^x e^i^y\]

If you take the modulus squared, you get:

\[e^2^x\]

and then square-rooting it gives just \[e^x\] .

So I know that x = 0, but what does y equal?

|e^z - 1|

if so ...

e^z - 1 = e^x*(cos(y) + i*sin(y)) - 1

so take the modulus of that where you have the real and imaginary parts.

Last edited by mqb2766; 1 year ago

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(Original post by

You plot an expression, not an equation. So you could plot

|e^z - 1|

if so ...

e^z - 1 = e^x*(cos(y) + i*sin(y)) - 1

so take the modulus of that where you have the real and imaginary parts.

**mqb2766**)You plot an expression, not an equation. So you could plot

|e^z - 1|

if so ...

e^z - 1 = e^x*(cos(y) + i*sin(y)) - 1

so take the modulus of that where you have the real and imaginary parts.

-taking the square root of (e^x*cos(y) - 1)^2 + (e^x*sin(y)) ?

If w = e^z, where w is a complex number (=r * e^iθ), then the above expression should give me r right?

But then how would I work out what theta is?

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#4

(Original post by

would taking the modulus would mean doing the following?:

-taking the square root of (e^x*cos(y) - 1)^2 + (e^x*sin(y)) ?

If w = e^z, where w is a complex number (=r * e^iθ), then the above expression should give me r right?

But then how would I work out what theta is?

**IDontKnowReally**)would taking the modulus would mean doing the following?:

-taking the square root of (e^x*cos(y) - 1)^2 + (e^x*sin(y)) ?

If w = e^z, where w is a complex number (=r * e^iθ), then the above expression should give me r right?

But then how would I work out what theta is?

z = x + iy, so the y-axis is your angle

Edit for theta, atan2(e^x*sin(y),e^x*cos(y) - 1)

Last edited by mqb2766; 1 year ago

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(Original post by

That is right for the modulus

z = x + iy, so the y-axis is your angle

Edit for theta, atan2(e^x*sin(y),e^x*cos(y) - 1)

**mqb2766**)That is right for the modulus

z = x + iy, so the y-axis is your angle

Edit for theta, atan2(e^x*sin(y),e^x*cos(y) - 1)

why can't you just write the modulus of e^z as e^x and equate it to 1, getting x = 0?

Because the modulus of e^iy = 1 for all y so x=0 for all y.

Why can't I just plot the x=0 line?

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#6

(Original post by

but here is what i dont understand:

why can't you just write the modulus of e^z as e^x and equate it to 1, getting x = 0?

Because the modulus of e^iy = 1 for all y so x=0 for all y.

Why can't I just plot the x=0 line?

**IDontKnowReally**)but here is what i dont understand:

why can't you just write the modulus of e^z as e^x and equate it to 1, getting x = 0?

Because the modulus of e^iy = 1 for all y so x=0 for all y.

Why can't I just plot the x=0 line?

As you say,

|e^z| = e^x

arg(e^z) = y

if you want to solve (not plot)

e^z = 1

you have to have y = k*2*pi so that it is a positive, real number and x = 0 so that it hits 1. If y is any other value, the imaginary component of e^z is non-zero (or negative real).

If you wanted to solve

|e^z| = 1,

then this only depends on x (=0). Any value of y would do.

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(Original post by

In general, if z = x+iy, whatever function or expression (modulus, argument, ..) you're plotting is a 3D surface on x & y.

As you say,

|e^z| = e^x

arg(e^z) = y

if you want to solve (not plot)

e^z = 1

you have to have y = k*2*pi so that it is a positive, real number and x = 0 so that it hits 1. If y is any other value, the imaginary component of e^z is non-zero (or negative real).

If you wanted to solve

|e^z| = 1,

then this only depends on x (=0). Any value of y would do.

**mqb2766**)In general, if z = x+iy, whatever function or expression (modulus, argument, ..) you're plotting is a 3D surface on x & y.

As you say,

|e^z| = e^x

arg(e^z) = y

if you want to solve (not plot)

e^z = 1

you have to have y = k*2*pi so that it is a positive, real number and x = 0 so that it hits 1. If y is any other value, the imaginary component of e^z is non-zero (or negative real).

If you wanted to solve

|e^z| = 1,

then this only depends on x (=0). Any value of y would do.

And if k can be any number, then k=0 would give y = 0, so would one of the points be a line of length 1 lying on the positive real axis?

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#8

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Okay so if you write the modulus of e^z as e^x*e^iy, its the same form as r* e^iθ, right? where r is e^x = 1 and y = θ = 2*k*pi?

And if k can be any number, then k=0 would give y = 0, so would one of the points be a line of length 1 lying on the positive real axis?

**IDontKnowReally**)Okay so if you write the modulus of e^z as e^x*e^iy, its the same form as r* e^iθ, right? where r is e^x = 1 and y = θ = 2*k*pi?

And if k can be any number, then k=0 would give y = 0, so would one of the points be a line of length 1 lying on the positive real axis?

e^z = e^x*e^iy which is now in polar form re^iθ, where

r = e^x = 1

y = θ = 2*k*pi

Yes, for the k = 0 point. In x-y space the exponential solutions repeat themselves every 2pi. So you can imagine horizontal bands of height 2pi, where the solution repeats itself for different values of k.

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(Original post by

Yes ...

e^z = e^x*e^iy which is now in polar form re^iθ, where

r = e^x = 1

y = θ = 2*k*pi

Yes, for the k = 0 point. In x-y space the exponential solutions repeat themselves every 2pi. So you can imagine horizontal bands of height 2pi, where the solution repeats itself for different values of k.

**mqb2766**)Yes ...

e^z = e^x*e^iy which is now in polar form re^iθ, where

r = e^x = 1

y = θ = 2*k*pi

Yes, for the k = 0 point. In x-y space the exponential solutions repeat themselves every 2pi. So you can imagine horizontal bands of height 2pi, where the solution repeats itself for different values of k.

Is the x-y space the same as Im(z)-Re(z)?

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#10

(Original post by

right, but convention is that -pi < arg(z) < pi, so would y = 0, y = 2pi, y = 4pi, etc just take you back to the same point, if e^x = 1 stays constant?

Is the x-y space the same as Im(z)-Re(z)?

**IDontKnowReally**)right, but convention is that -pi < arg(z) < pi, so would y = 0, y = 2pi, y = 4pi, etc just take you back to the same point, if e^x = 1 stays constant?

Is the x-y space the same as Im(z)-Re(z)?

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(Original post by

yes, yes (x-y is re(z)-im(z))

**mqb2766**)yes, yes (x-y is re(z)-im(z))

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#12

(Original post by

so why would there be horizontal bands of height 2pi where the solution repeats itself?

**IDontKnowReally**)so why would there be horizontal bands of height 2pi where the solution repeats itself?

y = 3pi

would be the same as

y = pi

would be the same as

y = 5pi ...

You're just wrapping round the circle again and again and ...

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(Original post by

because

y = 3pi

would be the same as

y = pi

would be the same as

y = 5pi ...

You're just wrapping round the circle again and again and ...

**mqb2766**)because

y = 3pi

would be the same as

y = pi

would be the same as

y = 5pi ...

You're just wrapping round the circle again and again and ...

why would you have an odd number of pi if y = 2*k*pi?

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#14

(Original post by

to be completely honest, im lost.

why would you have an odd number of pi if y = 2*k*pi?

**IDontKnowReally**)to be completely honest, im lost.

why would you have an odd number of pi if y = 2*k*pi?

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(Original post by

I just picked an angle "pi" at random. Pick which ever angle you want and add multples of 2pi to it.

**mqb2766**)I just picked an angle "pi" at random. Pick which ever angle you want and add multples of 2pi to it.

Thank you so much for all your help!

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