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Question

If z=1+cos6π5+isin6π5 , then

A
amp(z)=2π5
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B
amp(z)=2π5
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C
|z|=2cos2π5
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D
|z|=2cos2π5
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Solution

The correct option is C |z|=2cos2π5
z=1+cos6π5+isin6π5
|z|=(1+cos6π5)2+(sin6π5)2
=2+2cos6π5
=2(1+cos6π5)
=22cos2(3π5)
=2cos(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=sin6π51+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=1cosπ5isinπ5
z=2sin2π10i2sinπ10cosπ10
z=2sinπ10(sinπ10icosπ10)
z=2cos(π2π10)[cos(π2π10)isin(π2π10)]
z=2cos2π5[cos2π5isin2π5]
z=2cos2π5[cos(2π5)+isin(2π5)]
|z|=2cos2π5 and amp(z)=2π5

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