# Image of a line

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#1
Hi guys!

I have a question, if a complex function corresponds to a positive rotation about the point i by pi/4 radians. I get that the overall transformation is
(z-i)*(e^pi/4*i)+i

But I don't understand how to find the image of the line arg(z)=0 under f(z)?

What method can I use to do it algebraically?
I'm getting the answer as Re(z)

But honestly not sure with what I'm doing!

If anyone could help that you be great!
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1 month ago
#2
I think arg(z)=0 means z is a positive real number, and Im(z)=0, so instead of the usual z=x+yi, z=x in this case. Then you should be able to simplify it and match it to u+iv.
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#3
I think arg(z)=0 means z is a positive real number, and Im(z)=0, so instead of the usual z=x+yi, z=x in this case. Then you should be able to simplify it and match it to u+iv.
Thank you!
Ah so are you saying I should substitute z=x into this?
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1 month ago
#4
Yeah, and I expect after simplifying the e bit and separating the whole thing into the real and imaginary parts you'll have something to work with
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#5
Yeah, and I expect after simplifying the e bit and separating the whole thing into the real and imaginary parts you'll have something to work with
Well I tried expanding it in terms of cosine and sine
Getting ( root2/2+root2/2i)(x-i)+i
I have no clue what to do next after simplifying that
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1 month ago
#6
I'd expand and match the real parts to u (real part of w) and match the imaginary parts to v (imaginary part of w) and see if I can form some relationship.
1
#7
I'd expand and match the real parts to u (real part of w) and match the imaginary parts to v (imaginary part of w) and see if I can form some relationship.
I see I'll try this now! THANK YOU!
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#8
I'd expand and match the real parts to u (real part of w) and match the imaginary parts to v (imaginary part of w) and see if I can form some relationship.
I have tried this and get u=root2/2(x+y) and w= root2/2(x-y)+1
However I am not sure what to do next
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1 month ago
#9
(Original post by maths4life2020)
Hi guys!

I have a question, if a complex function corresponds to a positive rotation about the point i by pi/4 radians. I get that the overall transformation is
(z-i)*(e^pi/4*i)+i

But I don't understand how to find the image of the line arg(z)=0 under f(z)?

What method can I use to do it algebraically?
I'm getting the answer as Re(z)

But honestly not sure with what I'm doing!

If anyone could help that you be great!
Can you not just use the transformation directly
z = x*e^i0
So
(z-i)*(e^pi/4*i)+i = z*(e^pi/4*i) + i-ie^pi/4*i =
x*e^pi/4*i + offset
1
#10
(Original post by mqb2766)
Can you not just use the transformation directly
z = x*e^i0
So
(z-i)*(e^pi/4*i)+i = z*(e^pi/4*i) + i-ie^pi/4*i =
x*e^pi/4*i + offset
Oh thank you, I see. But how would I get the image of the line from this?
Sorry this question I've been working on all day so I have been going round in circles
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1 month ago
#11
(Original post by maths4life2020)
Oh thank you, I see. But how would I get the image of the line from this?
Sorry this question I've been working on all day so I have been going round in circles
x is the non-negative modulus (real component of original line). The transformed line is
offset + half line at 45 degrees
So just work out the offset (rotation of the orign).
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#12
(Original post by mqb2766)
x is the non-negative modulus (real component of original line). The transformed line is
offset + half line at 45 degrees
So just work out the offset (rotation of the orign).
Would it be pi/4?
As working it out to get i=ie^pi/4i
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1 month ago
#13
(Original post by maths4life2020)
Would it be pi/4?
As working it out to get i=ie^pi/4i
Would what be pi/4? You can do simple geometry to get the rotation of the origin by 45, or use the expression a couple of posts ago. It should be straightforward either way.
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#14
(Original post by mqb2766)
Would what be pi/4? You can do simple geometry to get the rotation of the origin by 45, or use the expression a couple of posts ago. It should be straightforward either way.
The angle of rotation?
Ah I'll keep on working on it!
Thanks for the help
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1 month ago
#15
(Original post by maths4life2020)
The angle of rotation?
Ah I'll keep on working on it!
Thanks for the help
Not sure what you don't understand?
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#16
(Original post by mqb2766)
Not sure what you don't understand?
I just don't get how to get the image of the line when arg(z)=0
My answers have been all over the place hahaha
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1 month ago
#17
(Original post by maths4life2020)
I just don't get how to get the image of the line when arg(z)=0
My answers have been all over the place hahaha
Copied from #9,

Can you not just use the transformation directly
z = x*e^i0
So
(z-i)*(e^pi/4*i)+i = z*(e^pi/4*i) + i-ie^pi/4*i = x*e^pi/4*i + offset

Which part don't you understand? I've left the offset (rotation of origin) for you to calculate.
0
#18
(Original post by mqb2766)
Copied from #9,

Can you not just use the transformation directly
z = x*e^i0
So
(z-i)*(e^pi/4*i)+i = z*(e^pi/4*i) + i-ie^pi/4*i = x*e^pi/4*i + offset

Which part don't you understand? I've left the offset (rotation of origin) for you to calculate.
From calculating the offset
I get ie^pi/4i=i
Do I then solve this?
To get -1+root2
0
1 month ago
#19
(Original post by maths4life2020)
From calculating the offset
I get ie^pi/4i=i
Do I then solve this?
To get -1+root2
ie^pi/4i=i
How?

ie^pi/4i = e^i3pi/4
As i = e^ipi/2

If you're unsure, just do some simple geometry.
Last edited by mqb2766; 1 month ago
0
#20
(Original post by mqb2766)
ie^pi/4i=i
How?

ie^pi/4i = e^i3pi/4
As i = e^ipi/2
Oh!!!! I see!! Thank you so much! So the image of the line is e^3pi/4i
I completely forgot about i = e^ipi/2
My brain wasn't working hahahaha
0
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