If -1- i -1- i - a - ib then show that a 2- b 2-1

The correct option is We have (a+ib)= √(1+i/1 i) =√(1+i)/1 i×√(1+i)/√(1+i) = (1+i)/√(1 i^2) =(1+i)/√(2) = ( 1/√(2)+1/√(2) i ) ⇒ |a+ib|^2= ( 1/√(2) )^2+ ...

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Byju's Answer

Standard XI

Mathematics

Inequality

If √1+ i /1- ...

Question

If $\sqrt{\frac{1 + i}{1 - i}} = (a + i b)$ then show that $(a^{2} + b^{2}) = 1$ .

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Solution

We have

$(a + i b) = \sqrt{\frac{1 + i}{1 - i}} = \frac{\sqrt{1 + i}}{1 - i} \times \frac{\sqrt{1 + i}}{\sqrt{1 + i}} = \frac{(1 + i)}{\sqrt{1 - i^{2}}}$

$= \frac{(1 + i)}{\sqrt{2}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i)$

$\Rightarrow | a + i b |^{2} = {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = (\frac{1}{2} + \frac{1}{2}) = 1.$

$\Rightarrow (a^{2} + b^{2}) = 1$ .

Hence, $(a^{2} + b^{2}) = 1.$

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If √1+i1−i=(a+ib) then show that (a2+b2)=1.

We have (a+ib)=√1+i1−i=√1+i1−i×√1+i√1+i=(1+i)√1−i2 =(1+i)√2=(1√2+1√2i) ⇒|a+ib|2=(1√2)2+(1√2)2=(12+12)=1. ⇒(a2+b2)=1. Hence, (a2+b2)=1.

If $\sqrt{\frac{1 + i}{1 - i}} = (a + i b)$ then show that $(a^{2} + b^{2}) = 1$ .