Question

(1+i)(1-i) power is -1 state the value of a and b

Answer 1

Given Expression,

$\sf{(1 + i)(1 - i) {}^{ - 1} } \\ \\ \ \implies \: \underline{\boxed{\sf{ \frac{1 + i}{1 - i} }}}$

Now,

$\implies \sf{\frac{1 + i}{1 - i} \: \times \: \frac{1 + i}{1 + i} } \\ \\ \implies \: \sf{ \frac{(1 + i) {}^{2} }{(1 + i)(1 - i)} } \\ \\ \implies \: \sf{ \frac{1 + i {}^{2} + 2i }{1 {}^{2} - i {}^{2} } } \\ \\ \sf{since \: i \: = - 1} \\ \\ \implies \: \sf{ \frac{2i}{2} } \: \implies \: \sf{i} \\ \\ \implies \: \huge{ \sf{0 + i}}$

•On comparing with z = a + bi,we get:

a = 0 and b = 1

Answer 2

Answer: a = 0 , b = 1

Step-by-step explanation:

Given  complex  number,

$(1+i)(1-i)^{-1}\\\;\\=\dfrac{1+i}{1-i}\\\;\\\text{Multiplying by (1+i) in denominator and numerator,}\\\;\\=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\\\;\\=\dfrac{(1+i)^2}{1^2-i^2}\\\;\\=\dfrac{1^2+i^2+2i}{1-(-1)}\\\;\\=\dfrac{1-1+2i}{1+1}\\\;\\=\dfrac{2i}{2}\\\;\\=i\\\;\\=0+i$

Comparing this complex number with  standard form of complex number a+ib;

We have,

a = 0

b = 1

Note: 1) i  = √(-1)

         2) i² = -1

         3) i³ = -i

         4) i⁴ = 1