If {left( {frac{{1 - i}}{{1 + i}}} right)^{100}} = a + ib , then

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4-1.Complex numbers

easy

If ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$, then

$a = 2,b = - 1$

$a = 1,b = 0$

$a = 0,b = 1$

$a = - 1,b = 2$

Solution

(b) Given, ${\left( {\frac{{1 – i}}{{1 + i}}} \right)^{100}} = a + ib$; $\left[ {\left( {\frac{{1 – i}}{{1 + i}}} \right) \times \left( {\frac{{1 – i}}{{1 – i}}} \right)} \right] = a + ib$
==> $a + ib = {\left[ {\frac{{{{(1 – i)}^2}}}{2}} \right]^{100}} = {\left[ {\frac{{ – 2i}}{2}} \right]^{100}} = {( – i)^{100}}$
==> $a + ib = {\left[ {{{(i)}^4}} \right]^{25}} = 1 + 0i,$Hence, $a = 1,b = 0$.

Standard 11

Mathematics

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medium

If ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$, then

Solution

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