Question

Convert z = sqrt 3-i in the polar form​

Answer 1

Given: z = √3 - i

To find: z in the polar form​.

Solution:

Let √3 be equal to (r cos θ) and the coefficient of i that is -1 be equal to (r sin θ). The two equations are squared. This can be represented as follows.

$r cos \theta = \sqrt{3}$

$r^{2} cos^{2} \theta = (\sqrt{3})^{2}$

$r sin \theta = -1$

$r^{2} sin^{2} \theta = (-1)^{2}$

Now, the two equations are added.

$r^{2} cos^{2} \theta + r^{2} sin^{2} \theta = (\sqrt{3})^{2} + (-1)^{2}$

$r^{2} (cos^{2} \theta + sin^{2} \theta) = 3 + 1$

$r^{2} = \sqrt{4}$

The sum of the squares of the sine and the cosine of an angle is equal to one. So, the value of r is 2. Now, the two equations can be written as follows.

$2 cos \theta = \sqrt{3}$

$cos \theta = \frac{\sqrt{3} }{2}$

$2 sin \theta = -1$

$sin \theta = \frac{-1}{2}$

As evident from the two equations, the value of theta is π/6 and -π/6, respectively. Hence, the polar form can be written as follows.

$\sqrt{3} - i = r cos \theta + i rsin \theta$

           $= 2 cos\frac{\pi }{6} - i2 sin \frac{\pi }{6}$

           $= 2 (cos\frac{\pi }{6} - i sin \frac{\pi }{6})$

Therefore, z in the polar form​ is 2(cos π/6 - i sin π/6).

Answer 2

z = √3 - i in the polar form is 2 cos(-π/6) + isin(-π/6)

Given:

• z = √3 - i

To Find:

• Polar form

Solution:

• The polar form of a complex number z=a+bi is z=r(cosθ+isinθ)  
• r=|z|=√a²+b² , a=rcosθ , b=rsinθ ,
• θ=tan⁻¹(b/a) for a>0 and θ=tan−1(b/a)+π for a<0

Step 1:

Compare z = √3 - i with z = a + bi

a = √3

b = - 1

Step 2:

Compare z = √3 - i with z = a + bi

a = √3

b = - 1

Step 3:

Calculate r=|z|=√a²+b²

r = √(√3)² + (-1)²

r = √4

r = 2

-ve value not considered as r is always +ve

Step 4:

Calculate θ=tan⁻¹(b/a) as  a>0

θ = tan⁻¹(-1/√3)

tanθ = -1/√3

θ = -π/6

Step 5:

Write in polar form

z= 2 cos(-π/6) + isin(-π/6)

z = √3 - i in the polar form is 2 cos(-π/6) + isin(-π/6)