Question

3, Iii

Answer 1

The given complex number is

1−i.

Let, rcosθ=1(1)

and rsinθ=−1(2)

Square and add equation (1) and equation (2).

( rcosθ ) 2 + ( rsinθ ) 2 = ( 1 ) 2 + ( −1 ) 2 r 2 ( cos 2 θ+ sin 2 θ )=1+1 r 2 =2 r=± 2

Since modulus is always positive, therefore take positive value of r.

The value of modulus of the complex variable is 2 .

Substitute 2 for r in equation (1).

2 cosθ=1 cosθ= 1 2 θ= π 4

Substitute 2 for r in equation (2).

2 sinθ=−1 sinθ=− 1 2 θ=− π 4

As, cosθ is positive and sinθ is negative, therefore 'θ' lies in the fourth quadrant.

So, the value of argument, that is θ is − π 4 .

The conversion of the complex number in polar form is,

z=r( cosθ+isinθ )

Substitute the values of z, r and θ in the above formula.

1−i= 2 [ cos( − π 4 )+isin( − π 4 ) ]

Thus, the complex number 1−i in the polar form is 2 [ cos( − π 4 )+isin( − π 4 ) ].