find the modulus & argument of the complex number 1+i
Answers
Answer:
Step-by-step explanation:
ANSWER
Let z=
1−i
1+i
Rationalizing the same,
z=
1−i
1+i
∗
1+i
1+i
z=
(1−i)(1+i)
(1+i)(1+i)
Using(a−b)(a+b)=a
2
−b
2
=
1
2
−i
2
(1+i)
2
Using(a+b)
2=
a
2
+b
2
+2ab
=
1
2
−i
2
1
2
+i
2
+2i
Puttingi
2
=−1
=
1−(−1)
1
2
+(−1)+2i
=
2
2i
=i
=0+i
Hencez=(0+i)
To calculate modulus of z,
z=(0+i)
Complexnumberzisoftheformx+iy
Hencex=0andy=1
Modulus of z=
x
2
+y
2
z=
0
2
+1
2
z=
0+1
z=
1
To find the argument,
0+i=rcosθ+irsinθ
Comparing real part,
0=rcosθ
Put r=1
0=1∗cosθ
0=cosθ
cosθ=0
Comparing imaginary part,
1=rsinθ
Put r=1
1=1∗sinθ
1=sinθ
sinθ=1
Hencecosθ=0andsinθ=1
Now , we have to find out argument of the complex number (1+i).
Answer :
Modulus of 1+i = √2
argument of 1+i = π/4
Additional Information :
Properties of Modules :-
If X = a+bi then ,
1) | X | = |- X |
2) | X₁ X₂ | = | X₁ | | X₂ |
3) | X₁ / X₂ | = | X₁ | / | X₂ |
4) | X₁ + X₂ | ≠ | X₁ | + | X₂ |
Properties of argument :-
1) Arg(0) = not defined
2) If X is purely imaginary number then , arg(X) = ± (π/2)
3) Arg( X₁ X₂) = Arg( X₁ ) + Arg( X₂) + 2mπ
4) Arg( X₁ - X₂) = Arg( X₁ ) - Arg( X₂) + 2mπ
5) Arg( Xⁿ) = n Arg( Xⁿ) + 2mπ