Let us simplify and express in the standard form of (a + ib),
\(\dfrac{ (1 – i)^3 }{ (1 – i^3) }\)
=\(\dfrac{ [1 – 3i + 3(-1)-i^2.i] }{ (1 – (-1)i)}\) [since, i2 = -1]
= \( \dfrac{-2-4i }{ (1+i)}\)
[Multiply and divide with (1-i)]
\( = \dfrac{-2-4i}{ (1+i)} × \dfrac{(1-i)}{(1-i)}\)
=\(\dfrac{ -2-2i+4(-1) }{2}\)
= \(\dfrac{ (-6-2i)}{2}\)
= -3 – i
The values of a, b are -3, -1
Answered by Aaryan | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
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