Complex Conjugate of this

Question

If $z$ is a complex number $a + bi$ and $z'$ is the conjugate $a - bi$, then would the conjugate of $(z + i)$, or $(z + i)'$, be $z - i$, or would it be $z' + i'$?

Answer 1

Complex conjugation is distributive over addition:

$ (z+i)'= {z}'+ {i}'=z'-i$.

Answer 2

$$\overline{(a + b i) + i} = \overline{a + (b+1) i} = a - (b+1)i$$

Answer 3

The complex conjugate of any real number (including the real part of any complex number) is itself.

Thusly taking $z=x+iy$ and $\hat{z}=\hat x+i\hat y$; we have that $$z'+\hat z '=x-iy+\hat x-i\hat y=(x+\hat x)-i(y+\hat y)=(z+\hat z)'$$

In your question, you simply have that $\hat z=i ( \text{i.e. }\hat x=0, \hat y=1)$

Answer 4

Since $z=a+ib$, you can write $(z+i)'=(a+ib+i)'=(a+i(b+1))'=a-i(b+1)=a-ib-i$. Now remember $z'=a-ib$ and $(i)'=-i$

From this you have $(z+i)'=a-ib-i=z'+(i)'$.

Please ask if something is unclear.