factoraisation
(1+a) ² (1+6)= (1+b)²(1ta)?
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Answer:
a+ib/1+c
Step-by-step explanation:
1+ca+ib
1−iz1+iz=1−i(b+ic)/(1+a)1+i(b+ic)/(1+a)
=1+a+c−ib1+a−c+ib
=(1+a+c)2+b2(1+a−c+ib)(1+a+c+ib)
=1+a2+c2+b2+2ac+2(a+c)1+2a+a2−b2−c2+2ib+2iab
=2+2ac+2(a+c)2a+2a2+2ib+2iab(∵a2+b2+c2=1)
=1+ac+(a+c)a+a2
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