Question

फाइंड द मॉडल्स एंड अरगुमेंट्स ऑफ ईच ऑफ द कंपलेक्स नंबर z=-root3 +iota​

Answer 1

Solution :

Here, z=−3–√+i

Comparing it with polar form, we get,

rcosθ=−3–√andrsinθ=1

Squaring and adding these two terms, we get,

r2cos2θ+r2sin2θ=4

r2(cos2θ+sin2θ)=4

r2=4⇒r=2

So, modulus is 2.

Now, rsinθ=1⇒sinθ=12

θ=π6

As cosθ is negative and sinθ is positive, theta lies in 2nd quadrant.

So, θ=π−π6=5π6