This is from a textbook:
$w=i\frac{1-z}{1+z}$ where $z$ is also a complex number
Then
$$\mathrm{Im}(w)=\frac{w-\bar w}{2i}$$
Then
$$\mathrm{Im}(w)=\frac 1 {2i} \left(i\frac{1-z}{1+z}+i\frac{1-\bar z}{1+\bar z}\right)$$
The last part is that part I don't understand. I have looked at several identities and I can't figure out why
$$-\bar w = i\frac{1-\bar z}{1+\bar z}$$
If someone could explain this to me I would appreciate it
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$\begingroup$This is simply because conjugation is a field homomorphism: if $w=\mathrm i\dfrac{1-z}{1+z}$, then $$\overline{w}=-\mathrm i\Bigl(\overline{\frac{1-z}{1+z}}\Bigr)=-\mathrm i\frac{\overline{1-z}}{\overline{1+z}}=-\mathrm i\frac{1-\bar z}{1+\bar z}.$$
$\endgroup$ 1 $\begingroup$$$\bar w = \bar{i}\frac{1-\bar z}{1+\bar z}$$
Note that $\bar{i}=-i$. So you get:
$$-\bar w = i\frac{1-\bar z}{1+\bar z}$$
$\endgroup$ $\begingroup$The complex conjugate distributes through addition and multiplication, so $\overline{(z+w)} = \bar z+\bar w$ and $\overline{(zw)} = \bar z\bar w$. Division is just multiplication by the reciprocal, and the conjugate distributes through that too, so $\overline{(z/w)} = \bar z/\bar w$. Finally, the distinctive property of the conjugate is that it maintains the real part while it flips the sign of the imaginary part, so that $\bar 1 = 1$ and $\bar i = -i$. Put all of these rules together and you can see why the relation you are wondering about must be true.
$\endgroup$ $\begingroup$You should understand it in the following way: $$w=\frac{a+b z}{c+dz}=\frac{(a+b z)(c^*+d^* z^*)}{(c+dz)(c^*+d^* z^*)}$$ The Denominator is a real number thus $$ w^* = \frac{(a^*+ b^* z^*)(c+d z)}{(c+dz)(c^*+d^* z^*)}=\frac{(a^*+ b^* z^*)}{c^*+d^* z^*} $$ I use the following identity if $z_1$ and $z_2$ are complex number $(z_1 z_2)^*=z_1^* z_2^*$ and $z_1^*$ is a complex conjugate number.
$\endgroup$ $\begingroup$We will use the following two theorems.
Theorem 1:
The complex conjugate of the sum (difference) of two complex numbers is the sum (difference) of the complex conjugates.
Proof: To show this, let $z_1=x_1+y_1$ and $z_2=x_2+y_2$. Then,
$$\begin{align} \overline{\left(z_1\pm z_2\right)}&=\overline{(x_1\pm x_2)+i(y_1\pm y_2)}\\\\ &=(x_1\pm x_2)- i(y_1\pm y_2)=(x_1-iy_1)\pm (x_2-iy_2)\\\\ &=\overline{z_1}\pm \overline{z_2} \end{align}$$
Theorem 2:
The complex conjugate of the product (quotient) of two complex numbers is the product (quotient) of the complex conjugates.
Proof: To show this for the product, we again let $z_1=x_1+y_1$ and $z_2=x_2+y_2$. Then,
$$\begin{align} \overline{z_1z_2}&=\overline{(x_1+iy_1)(x_2+iy_2)}\\\\ &=\overline{(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)}\\\\ &=(x_1x_2-y_1y_2)-i(x_1y_2+x_2y_1)\\\\ &=(x_1-iy_1)(x_2-iy_2)\\\\ &(\bar z_1)(\bar z_2) \end{align}$$
Proof: To show this for the quotient, we again let $z_1=x_1+y_1$ and $z_2=x_2+y_2$. Then,
$$\begin{align} \overline{z_1/z_2}&=\overline{(x_1+iy_1)/(x_2+iy_2)}\\\\ &=\overline{\frac{(x_1+iy_1)(x_2-iy_2)}{|z_2|^2}}\\\\ &=\overline{\frac{(x_1x_2+y_1y_2)-i(x_1y_2-x_2y_1)}{|z_2|^2}}\\\\ &=\frac{(x_1x_2+y_1y_2)+i(x_1y_2-x_2y_1)}{|z_2|^2}\\\\ &=\frac{(x_1-iy_1)(x_2+iy_2)}{|z_2|^2}\\\\ &=\frac{x_1-iy_1}{x_2-iy_2}\\\\ &=(\bar z_1)/(\bar z_2) \end{align}$$
Now, for the problem of interest, we have
$$w=i\left(\frac{1-z}{1+z}\right)$$
Thus, the complex conjugate of $w$ is
$$\begin{align} \bar w &=\overline{i\left(\frac{1-z}{1+z}\right)} \tag 1 \\\\ &=-i\overline{\left(\frac{1-z}{1+z}\right)} \tag 2 \\\\ &=-i\frac{\overline{1-z}}{\overline{1+z}} \tag 3 \\\\ &=-i\frac{1-\bar z}{1+\bar z}\tag 4 \end{align}$$
where in going from $(1)$ to $(2)$ we used the Theorem 2 for the product rule, in going $(2)$ to $(3)$ we used Theorem 2 for the quotient rule, and in going from $(3)$ to $(4)$ we used Theorem 1 for both the sum and the difference.
And we are done!
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