If z=\[\frac{a+ib}{c+id}\]
then after rationalizing we get =\[\frac{(a+ib)(c-id)}{(c+id)(c-id)}\]=\[\frac{ac+bd+i(bc-ad)}{(c^2+d^2)}\]
The conjugate of a complex number z=a+ib is denoted by \[\bar{z}\]and is defined as \[\bar{z}=a-ib\].
Case1 if Z= 2+3i then \[\bar{z}=2-3i\].
Case 2 If \[z_1=a+ib\] and \[z_2=c-id \]
then conjugate of \[z_1+z_2\]=conjugate of \[z_1\]+conjugate of \[z_2\] ..............................property -1
For Example \[z_1=5+6i\] and \[z_2=3-8i \] then conjugate of \[z_1+z_2\]=8+2i
Case 3 If \[z_1=a+ib\] and \[z_2=a-ib \]
then conjugate of \[z_1+z_2\]=conjugate of \[z_1\]+conjugate of \[z_2\]=\[2a\]....................special case
Note - \[z_1\] and \[z_2\] are cojugate of each other
Case 4. If imaginary part of complex number is zero then conjugate of the complex number is number itself.
Example z=2 then \[\bar{z}\]=2
Case 5. When real part of complex number is zero.
Example z=2i then \[\bar{z}\]=-2i
