Question

If ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$, then:
a) $a = 2,b = - 1$
b) $a = 1,b = 0$
c) $a = 0,b = 1$
d) $a = - 1,b = 2$

Answer 1

Hint: In the given problem, we are required to simplify an expression involving complex numbers. For simplifying the given expression, we need to have a thorough knowledge of complex number sets and its applications in such questions. Algebraic rules and properties also play a significant role in simplification of such expressions.

Formula used:
${i^2}=-1$
$\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$

Complete step by step solution:
 We will first simplify the expression $\left( {\dfrac{{1 - i}}{{1 + i}}} \right)$ and then find ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}}$ and equate the expression with $a + ib$ to find the values of a and b.
So, $\left( {\dfrac{{1 - i}}{{1 + i}}} \right) = \left( {\dfrac{{1 - i}}{{1 + i}}} \right) \times \left( {\dfrac{{1 - i}}{{1 - i}}} \right)$
Using the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$,
\[ \Rightarrow \left( {\dfrac{{1 - i}}{{1 + i}}} \right) = \dfrac{{{{\left( {1 - i} \right)}^2}}}{{{1^2} - {i^2}}}\]
We know that ${i^2} = - 1$. Hence, substituting ${i^2}$ as $ - 1$, we get,
\[ \Rightarrow \left( {\dfrac{{1 - i}}{{1 + i}}} \right) = \dfrac{{{{\left( 1 \right)}^2} - 2\left( i \right)\left( 1 \right) + {{\left( i \right)}^2}}}{{1 - \left( { - 1} \right)}}\]
Simplifying the expression, we get,
\[ \Rightarrow \left( {\dfrac{{1 - i}}{{1 + i}}} \right) = \dfrac{{1 - 2i - 1}}{{1 + 1}}\]
\[ \Rightarrow \left( {\dfrac{{1 - i}}{{1 + i}}} \right) = \dfrac{{ - 2i}}{2} = - i\]
We have, ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$
$ \Rightarrow {\left( { - i} \right)^{100}} = a + ib$
$ \Rightarrow {i^{100}} = a + ib$
Now, we know that ${i^4} = 1$.
$ \Rightarrow {\left( {{i^4}} \right)^{25}} = a + ib$
$ \Rightarrow 1 = a + ib$
Comparing both sides of equation, we get,
$a = 1,b = 0$
Hence, option (b) is the correct answer.

Note: The conjugate of a complex number is the complex number having negative imaginary part and same real part with respect to the original complex number. We must remember the algebraic identities such as $\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$ and ${\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right)$. We can compare the real and imaginary parts of the complex numbers, if they are given to be equal.