modulus and amplitude of
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Given z=−2+2
3
i
Hence, ∣z∣=2
1+3
⇒2
4
⇒2×2=4
And let α be the argument of the complex number.
Then
tanα=−
3
Now Re(z)<0 and Im(z)>0
∴ the number lies in the II quadrant
Hence,
α=π−
3
π
=
3
2π
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