Question

modulus and argument of -16/1+i√13 in complex number​

Answer 1

Answer:

z=−1−i 3

Let rcosθ=−1andrsinθ=−3

On squaring and adding, we obtain(rcosθ) 2. (rsinθ) 2 =(−1)

2+(− 3 ) 2⇒r 2

(cos 2 θ+sin 2 θ)=1+3⇒r 2 =4[cos

2θ+sin 2 θ=1]⇒r= 4

=2 (Conventionally,r>0 )

∴ Modulus of z i.e ∣z∣=2

∴2cosθ=−1and2sinθ=−

3

Since both the values of sinθ and cosθ are negative and sinθ and cosθ are negative in III quadrant,

Argument =−(π−

3

π

)=

3

−2π

Thus, the modules and argument of the complex number −1−

3i are 2 and 3 −2π respectively...

Thnx (￣ヘ￣)

Answer 2

Answer:

z=−1−i 3

Let rcosθ=−1andrsinθ=− 3

On squaring and adding, we obtain

(rcosθ)

2 +(rsinθ) 2=(−1)

2 +(− 3 ) 2⇒r

2 (cos 2 θ+sin 2θ)=1+3⇒r

2=4[cos 2 θ+sin 2 θ=1]⇒r

= 4

=2 (Conventionally,r>0 )

∴ Modulus of z i.e ∣z∣=2

∴2cosθ=−1and2sinθ=−

3

Since both the values of sinθ and cosθ are negative and sinθ and cosθ are negative in III quadrant,

Argument =−(π−

3

π

)=

3

−2π

Thus, the modules and argument of the complex number −1−

3i are 2 and 3 −2π respectively.