Thread: Complex Analysis (A Few Clarifying Questions On Practice Problems)

1. Complex Analysis (A Few Clarifying Questions On Practice Problems)

I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

1) Find a parameterization of the line segment joining $\small 2i$ and $\small 1+i$. Find the image of the segment under the mapping $\small f(z)=i\bar{z}$.
This is what I did:
$\small z_{o} = 2i$ and $\small z_{1} = 1+i$
$\small z(t)=z_{0}(1+t)+z_{1}t$ for $\small 0\leq t\leq 1$
$\displaystyle =2i(1-t)+(1+t)i$
$\displaystyle =2i-2t+i-1$
$\displaystyle =3i-2it-1$
$\displaystyle f(z)=i(\bar{3i-2it-1})$
$\displaystyle =i(\bar{-1+3-2t)i})$
$\displaystyle =i(-1-(3-2t)i)$
$\displaystyle =i(-1-(3i-2it))$
$\displaystyle =i(-1-3i+2it)$
$\displaystyle =-i-3i^{2}+2i^{2}t$
$\displaystyle =-i+3-2t$ for $\displaystyle 0\leq t \leq 1$

The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

2)Find the image $\displaystyle S'$ of the set $\displaystyle S$ under the given complex mapping $\displaystyle w=f(z)$
$\displaystyle f(z)=\bar{z}$; $\displaystyle S$ is the line $\displaystyle y=x$

A friend told me to do it this way:
$\displaystyle Re(z)=Im(z)$ $\displaystyle \therefore$ $\displaystyle w=\bar{z}=Re(z)=-Im(z)$
Is this correct?

I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.

2. Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

Originally Posted by DrKittenPaws
I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

1) Find a parameterization of the line segment joining $\small 2i$ and $\small 1+i$. Find the image of the segment under the mapping $\small f(z)=i\bar{z}$.
This is what I did:
$\small z_{o} = 2i$ and $\small z_{1} = 1+i$
$\small z(t)=z_{0}(1+t)+z_{1}t$ for $\small 0\leq t\leq 1$
$\displaystyle =2i(1-t)+(1+t)i$
$\displaystyle =2i-2t+i-1$
$\displaystyle =3i-2it-1$
$\displaystyle f(z)=i(\bar{3i-2it-1})$
$\displaystyle =i(\bar{-1+3-2t)i})$
$\displaystyle =i(-1-(3-2t)i)$
$\displaystyle =i(-1-(3i-2it))$
$\displaystyle =i(-1-3i+2it)$
$\displaystyle =-i-3i^{2}+2i^{2}t$
$\displaystyle =-i+3-2t$ for $\displaystyle 0\leq t \leq 1$

The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

2)Find the image $\displaystyle S'$ of the set $\displaystyle S$ under the given complex mapping $\displaystyle w=f(z)$
$\displaystyle f(z)=\bar{z}$; $\displaystyle S$ is the line $\displaystyle y=x$

A friend told me to do it this way:
$\displaystyle Re(z)=Im(z)$ $\displaystyle \therefore$ $\displaystyle w=\bar{z}=Re(z)=-Im(z)$
Is this correct?

I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.

the parameterization of the segment is simply

$2\imath + (1-\imath)t~~0\leq t \leq 1$

try redoing the problem with this.

3. Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

Originally Posted by romsek
the parameterization of the segment is simply

$2\imath + (1-\imath)t~~0\leq t \leq 1$

try redoing the problem with this.
Well, that was dumb of me. Let me try to work it out again and I'll get back to ya on what I got. Thanks!

4. Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

Originally Posted by DrKittenPaws
I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

1) Find a parameterization of the line segment joining $\small 2i$ and $\small 1+i$. Find the image of the segment under the mapping $\small f(z)=i\bar{z}$.
This is what I did:
$\small z_{o} = 2i$ and $\small z_{1} = 1+i$
$\small z(t)=z_{0}(1+t)+z_{1}t$ for $\small 0\leq t\leq 1$
$\displaystyle =2i(1-t)+(1+t)i$
$\displaystyle =2i-2t+i-1$
$\displaystyle =3i-2it-1$
$\displaystyle f(z)=i(\bar{3i-2it-1})$
$\displaystyle =i(\bar{-1+3-2t)i})$
$\displaystyle =i(-1-(3-2t)i)$
$\displaystyle =i(-1-(3i-2it))$
$\displaystyle =i(-1-3i+2it)$
$\displaystyle =-i-3i^{2}+2i^{2}t$
$\displaystyle =-i+3-2t$ for $\displaystyle 0\leq t \leq 1$

The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

2)Find the image $\displaystyle S'$ of the set $\displaystyle S$ under the given complex mapping $\displaystyle w=f(z)$
$\displaystyle f(z)=\bar{z}$; $\displaystyle S$ is the line $\displaystyle y=x$

A friend told me to do it this way:
$\displaystyle Re(z)=Im(z)$ $\displaystyle \therefore$ $\displaystyle w=\bar{z}=Re(z)=-Im(z)$
Is this correct?

I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.

You should have $z_0(1 - t)$ and not $z_0(1 + t)$.

For your second problem, you have the elements on your line are complex numbers of the form: $z = x + ix$.

From this, we see that the images are of the form: $f(z) = \overline{z} = \overline{x+ix} = x - ix$, so that these lie on the line $y = -x$.

5. Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

Originally Posted by DrKittenPaws
I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.
1) Find a parameterization of the line segment joining $\small 2i$ and $\small 1+i$. Find the image of the segment under the mapping $\small f(z)=i\bar{z}$.
Consider a vector approach. Express the points a vectors.
$(0,2)~\&~(1,1)\\(0,2)+t(1,-1),~0\le t\le 1\\(t,2-t),~0\le t\le 1\\t+i(2-t),~0\le t\le 1$
that is the parameterzation of the line segment.