If b - ic -1- a z and a 2- b 2- c 2-1- then 1- i z-1- i z is equal toA- a-i b-1-c- a-i b-1-cC- a-i b-1-c1-c a-i b 1-c

The correct option is D We have, b + ic = (1 + a)z ⇒ z = b + ic/1 + a⇒ iz = ib c/1 + a 1+iz/1 iz= (1+a c)+ib/(1+a+c) ib×(1 + a + c)+ib/(1+a+c)+ib = a^2 + a ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If b + ic =1+...

Question

$I f b + i c = (1 + a) z a n d a^{2} + b^{2} + c^{2} = 1, t h e n \frac{1 + i z}{1 - i z} i s e q u a l t o$

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Solution

The correct option is D
$W e h a v e, b + i c = (1 + a) z$
$\Rightarrow z = \frac{b + i c}{1 + a} \Rightarrow i z = \frac{i b - c}{1 + a} \frac{1 + i z}{1 - i z} = \frac{(1 + a - c) + i b}{(1 + a + c) - i b} \times \frac{(1 + a + c) + i b}{(1 + a + c) + i b} = \frac{a^{2} + a + i a b + i b}{1 + a + a c + c} = \frac{(a + 1) (a + i b)}{(a + 1) (1 + c)} = \frac{a + i b}{1 + c}$

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If b+ic=(1+a)z and a2+b2+c2=1,then1+iz1−iz is equal to

The correct option is D We have,b+ic=(1+a)z ⇒z=b+ic1+a⇒iz=ib−c1+a1+iz1−iz=(1+a−c)+ib(1+a+c)−ib×(1+a+c)+ib(1+a+c)+ib=a2+a+iab+ib1+a+ac+c=(a+1)(a+ib)(a+1)(1+c)=a+ib1+c

$I f b + i c = (1 + a) z a n d a^{2} + b^{2} + c^{2} = 1, t h e n \frac{1 + i z}{1 - i z} i s e q u a l t o$