taking a congugate of this question.z=(a+ib/a-ib)hole power two+(a-ib/a+ib)hole power two.

Question

calculate the conjugate complex number of,z=(a+ib/a-ib)2+(a-ib/a+ib)2

Answer 1

((a+ib)/(a-ib))²=(a+ib)⁴/(a²+b²)²;

((a-ib)/(a+ib))²=(a-ib)⁴/(a²+b²)²;

add these together:

((a+ib)⁴+(a-ib)⁴)/(a²+b²)².

(a+ib)⁴=a⁴+4a³ib-6a²b²-4aib³+b⁴;

(a-ib)⁴=a⁴-4a³ib-6a²b²+4aib³+b⁴;

add these together:

2a⁴-12a²b²+2b⁴;

divide by (a²+b²)²:

2(a⁴-6a²b²+b⁴)/(a²+b²)², which has only real components, making its conjugate the same as itself.