Let z \(\in\) c and z=a+ib, then conjugate of z, denoted by \(\overline{z}\)=a-ib
Example: z=2-i, then \(\overline{z}\)=2+i
z\(\overline{z}\)=(2-i)(2+i)
=\(2^{2}-(i)^{2}\)
=4+1=5 \(\in\) R
Conjugate of -3i+2 is 3i+2
Conjugate of -5-3i is -5+3i
Properties of conjugate:
If \(z_{1}\) and \(z_{1}\) complex number such that
\(z_{1}\)=x+iy and \(z_{2}\)=a+ib where i=\(\sqrt{-i}\), then
(i) \(\overline{(\overline{z_{1}})}\)=\(z_{1}\)
Proof: \(\overline{(\overline{z_{1}})}\)=x+iy=\(z_{1}\)
(ii) \(\overline{z_{1}+z_{2}}\)=\(\overline{z_{1}}+\overline{z_{2}}\)
Proof: \(z_{1}\)+\(z_{2}\)=x+iy+a+ib=(x+a)+i(y+b)
Now, \(\overline{(\overline{z})}\)=(x+a)-i(y+b)
=(x-iy)+(a-ib)=\(\overline{z_{1}}+\overline{z_{2}}\)
(iii) \(\overline{(z_{1}z_{2})}=\overline{z_{1}}.\overline{z_{2}}\)
Proof: z_{1}z_{2}=(x+iy)(a+ib)=ax+ibx-by
=(ax-yb)+i(bx+ay)
Now, \(\overline{z_{1}}.\overline{z_{2}}\)=(x+iy)(a-ib)
=(ax-by)+i(ay+bx)
(iv) \(\overline{\Big(\frac{z_{1}}{z_{2}}\Big)}\)=\(\frac{\overline{z_{1}}}{\overline{z_2}}\) provide \(z_{2}\neq0\)
Let, \(\frac{z_{1}}{z_{2}}\)=\({z_{3}}\)
\(\Rightarrow\) \(z_{1}=z_{2}z_{3}\)
\(\Rightarrow\) \( \overline{z_{1}}=\overline{z_{2}}\overline{z_{3}}\)
\(\Rightarrow\) \(\frac{\overline{z_{1}}}{\overline{z_2}}\)=\(\overline{z_{3}}\)=\(\overline{\Big(\frac{z_{1}}{z_{2}}\Big)}\)