Question

find the modulus and arguments of 1/1+i

Answer 1

Let  a complex number (z) = 1/1+i
$z= \frac{1(1-i)}{(1+i)(1-i)} \\ z= \frac{1-i}{ 1^{2} - i^{2} } \\ z= \frac{1-i}{1-(-1)} = \frac{1-i}{1+1 } \\ z= \frac{1-i}{2}$
now,
z = 1/2 -i/2
 so real part of z = x = 1/2
Imaginary part of z = y = 1/2
∴$|z|= \sqrt{ x^{2} + y^{2} } \\ |z|= \sqrt{ (1/2)^{2} + (1/2)^{2} } \\ |z|= \sqrt{ \frac{2}{ 2^{2} } } = \frac{1}{ \sqrt{2} }$

Argument = tan⁻¹(y/x)
$Argument = tan^{-1} \frac{1/2}{1/2} = tan^{-1} \frac{1}{1} = \frac{ \pi }{4}$

Answer 2

Answer:

1/√2

Step-by-step explanation:

1st step by rationalization : we get (1-i)/2 = 1/2 - i/2

2nd step comparing 1/2 - i/2 with x+iy we get x=1/2 & y=1/2

3rd step put the value of x & y in √(x²+y²) and

1/√2 is obtained