The modulus of (3+1)(2-1)/1+i
find the value of i
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Answer:
¡ = 1/4
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1+i/1-i - 1-i/1+i
{Taking LCM of both sides}
=(1+i)^2 - (1-i)^2/(1-i)(1+i)
= 1+i^2+2i-1-i^2+2i / 1^2 -(i^2)
{Putting value of i^2}
= 1-1+2i-1+1+2i / 1+1
= 4i / 2
=0+2i
Modulus of complex no is underrot x^2+y^2: where x is real part and y is img part.
Z= underoot (0^2) + (2)^2
= underoot 0+4
= underoot 4
=2
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