Question

convert the complex number -1+I in the polar form
​

Answer 1

Answer:

Given, z=1−i

Let rcosθ=1andrsinθ=−1

On squaring and adding, we obtain

r

2

cos

2

θ+r

2

sin

2

θ=1

2

+(−1)

2

⇒r

2

(cos

2

θ+sin

2

θ)=2

⇒r

2

=2

⇒r=

2

(since,r>0 )

∴

2

cosθ=1 and

2

sinθ=−1

∴θ=−

4

π

(As θ lies in fourth quadrant.)

So, the polar form is

∴1−i=rcosθ+irsinθ=

2

cos(

4

−π

)+i

2

sin(

4

−π

)

=

2

[cos(

4

−π

)+isin(

4

−π

)]

Step-by-step explanation:

hope it will help you

Answer 2

Answer:

The polar form of a complex number is written in any of the following forms: rcos θ + irsin θ, r(cos θ + isin θ), or rcis θ. In any of these forms r is called the modulus or absolute value. θ is called the argument.