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Q : 15     Find the modulus of   \small \frac{1+i}{1-i}-\frac{1-i}{1+i}.

Answers (1)

best_answer

Let
z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i}
Now, we will reduce it into 
z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i} = \frac{(1+i)^2-(1-i)^2}{(1+i)(1-i)}= \frac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2}= \frac{4i}{1+1}= \frac{4i}{2}=2i
Now,
r\cos\theta = 0 \ \ and \ \ r\sin \theta = 2
square and add both the sides. we will get,
r^2\cos^2\theta+r^2\sin^2 \theta = 0^2+2^2
r^2(\cos^2\theta+\sin^2 \theta) = 4
r^2 = 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos^2\theta+\sin^2 \theta = 1)
r = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)

Therefore, modulus of 

 \small \frac{1+i}{1-i}-\frac{1-i}{1+i}  is   2

Posted by

Gautam harsolia

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