Question

S, 717i

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

The value of modulus of 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 negative and sinθ is also negative, therefore 'θ' lies in the third quadrant.

So, value of θ is −( π− π 4 )=− 3π 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 equation,

−1−i= 2 [ cos( −3π 4 )+isin( −3π 4 ) ]

Thus, the complex number −1−i in the polar form is 2 [ cos( −3π 4 )+isin( −3π 4 ) ].