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I have a difficulty understanding the basics of complex functions. My exercise looks like this:

"The $z$-plane region $D$ consists of the complex numbers $z = x + yi$ that satisfy the given conditions: $$x + y = 1, w = \bar{z}$$ Describe the image $R$ of $D$ in the $w$-plane under the given function $w = f(z)$."

I just really don't know how to tackle this exercise, I know it's basic but any suggestions on how to go at it would be appreciated.

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  • $\begingroup$ What is the function $f$? Do you mean $f(z) = \bar{z}$? $\endgroup$ – Empiricist Jan 19 '15 at 9:42
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    $\begingroup$ you can think of $\bar z$ geometrically as the image of $z$ on the mirror $x$-axis. image of a straight line is a straight line. $\endgroup$ – abel Jan 19 '15 at 9:45
  • $\begingroup$ @SRX The exercise is now written exactly as it was given in the textbook. $\endgroup$ – martin Jan 19 '15 at 9:49
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$$x+y=1\iff y=1-x$$

and you have the straight line $\;y=1-x\;$ in the complex plane, which you can also express as the set

$$\{z\in\Bbb C\;:\;\;z=x+(1-x)i\;,\;\;x\in\Bbb R\}$$

If you take a general element of this set and apply on it the transformation $\;w\;$ ,we get

$$w(x+(1-x)i):=x-(1-x)i$$

and the image is the straight line $\;y=-(1-x)=x-1\;$.

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the set $D = \{x + iy: x + y = 1\}$ in the $z$ place is a line through $1$ and $i.$

the function $w = \bar z$ maps the point $1$ to itself and $i$ to $-i$. so the image of $D$ in the $w$ plane is a line through $1$ and $-i$ that is $R =\{\xi + i\eta \colon \xi - \eta = 1 \}$.

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