If a + ib = (1-I) ^ 3/ 1-i ^ 3 ,find a and b.
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2
Answer:
i^2 = - 1
a+ib= (1-i)^3/1+i)=[1-3i+3i^2 - i^3] /(1+i)
=[1-3i-3+i]/(1+i)
=-2(1+i)/(1+i) =-2
a=-2,b=0
Answered by
1
Step-by-step explanation:
a+ib = (1-i)^3/(1-i^3)
=(1-i)^3/(1-i)(1+i+i^2) [a^3-b^3 = (a-b)(a^2+b^2+ab)]
=(1-i)^2/(1-1+i) [i^2= -1]
= 1-1-2i/i
= -2i/i
a+ib = -2
On comparing a and b,
a= -2, b= 0
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