# Graphs of complex functions

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#1
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?
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1 year ago
#2
(Original post by 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?
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.
Last edited by mqb2766; 1 year ago
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#3
(Original post by 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.
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?
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1 year ago
#4
(Original post by 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?
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)
Last edited by mqb2766; 1 year ago
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#5
(Original post by 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)
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?
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1 year ago
#6
(Original post by 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?
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.
1
#7
(Original post by 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.
Okay so if you write the modulus of e^x 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?
0
1 year ago
#8
(Original post by 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?
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.
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#9
(Original post by 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.
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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1 year ago
#10
(Original post by 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)?
yes, yes (x-y is re(z)-im(z))
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#11
(Original post by mqb2766)
yes, yes (x-y is re(z)-im(z))
so why would there be horizontal bands of height 2pi where the solution repeats itself?
0
1 year ago
#12
(Original post by IDontKnowReally)
so why would there be horizontal bands of height 2pi where the solution repeats itself?
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 ...
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#13
(Original post by 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 ...
to be completely honest, im lost.
why would you have an odd number of pi if y = 2*k*pi?
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1 year ago
#14
(Original post by IDontKnowReally)
to be completely honest, im lost.
why would you have an odd number of pi if y = 2*k*pi?
I just picked an angle "pi" at random. Pick which ever angle you want and add multples of 2pi to it.
1
#15
(Original post by mqb2766)
I just picked an angle "pi" at random. Pick which ever angle you want and add multples of 2pi to it.
I see, so y = theta + 2pi*k?

Thank you so much for all your help!
1
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