For all z C, prove that

Solution:

Let z = a + ib

$\Rightarrow \bar{z}=a-i b$

Now,$\frac{z+\bar{z}}{2}=\frac{(a+i b)+(a-i b)}{2}=\frac{2 a}{2}=a=\operatorname{Re}(z)$

Hence Proved.

(ii) Let $z=a+i b$

$\Rightarrow \bar{z}=a-i b$

$w, \frac{z+\bar{z}}{2}$

$=\frac{(a+i b)+(a-i b)}{2}$

$=\frac{2 a}{2}=\frac{a}{1}=\operatorname{Re}(z)$

Hence, Proved.

(iii) Let $z=a+i b$

$\Rightarrow \bar{z}=a-i b$

Now, $z \bar{z}=(a+i b)(a-i b)=a^{2}-(i b)^{2}=a^{2}+b^{2}=|z|^{2}$

Hence Proved

(iv) Let $z=a+i b$

$\Rightarrow \bar{z}=a-i b$

Now,$z+\bar{z}=(a+i b)+(a-i b)=2 a=2 \operatorname{Re}(z)$

Hence, $^{Z}+\bar{z}$ is real.

(v) Case 1 . Let $z=a+0 i$

$\Rightarrow \bar{z}=a-0 i$

Now, $z-\bar{z}=(a+0 \mathrm{i})-(a-0 \mathrm{i})=0$

Case 2 . Let $z=0+b i$

$\Rightarrow \bar{z}=0-b i$

Now, $z-\bar{z}=(0+i b)-(0-i b)=2 i b=2 i \operatorname{Im}(z)=$ Imaginary

Case 2 . Let $z=a+i b$

$\Rightarrow \bar{z}=a-i b$

Now,$z-\bar{z}=(a+i b)-(a-i b)=2 i b=2 i \operatorname{Im}(z)=$ Imaginary

Thus, $(z-\bar{z})$ is 0 or imaginary.