The correct option is C |z|=2cos2π5
z=1+cos6π5+isin6π5
|z|=√(1+cos6π5)2+(sin6π5)2
=√2+2cos6π5
=√2(1+cos6π5)
=√2⋅2cos2(3π5)
=2∣∣∣cos(3π5)∣∣∣
=−2cos(3π5) [∵cos3π5<0]
=2cos(2π5)
Re(z)=1+cos6π5 >0
Im(z)=sin6π5 <0
So, z lies in 4th quadrant of Argand plane
tanα=∣∣∣sin6π5∣∣∣∣∣∣1+cos6π5∣∣∣
=∣∣
∣
∣
∣∣2sin3π5cos3π52cos23π5∣∣
∣
∣
∣∣
=∣∣∣tan3π5∣∣∣
=−tan3π5=tan2π5
⇒α=2π5
Now, amp(z)=θ=−α=−2π5
Alternate Solution:
z=1+cos6π5+isin6π5
⇒z=1+cos(π+π5)+isin(π+π5)
⇒z=1−cosπ5−isinπ5
⇒z=2sin2π10−i2sinπ10cosπ10
⇒z=2sinπ10(sinπ10−icosπ10)
⇒z=2cos(π2−π10)[cos(π2−π10)−isin(π2−π10)]
⇒z=2cos2π5[cos2π5−isin2π5]
⇒z=2cos2π5[cos(−2π5)+isin(−2π5)]
∴|z|=2cos2π5 and amp(z)=−2π5