Question

represent following complex no. in polar form 1-√3i​

Answer 1

Answer

Let z=−1−

3

i

Since −1−

3

i lies in third quadrant.

∴ Principal value of argz=−π+tan

−1

∣

∣

∣

∣

∣

∣

−1

−

3

∣

∣

∣

∣

∣

∣

−π+tan

−1

3

=−π+

3

π

=

3

−3π+π

=

3

−2π

and ∣z∣=

(−1)

2

+(−

3

)

2

=

1+3

=2

∴ Polar form of z=∣z∣[cos(argz)+isin(argz)]

Polar form of z=2[cos(

3

−2π

)+isin(

3

−2π

)]

Answer 2

Answer:

r(cos theeta - iota sin theeta)=2(1/2-root3/2)=2(cos60-iota sin60)

so theeta =60