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Question

If arg(z)=π3 and arg(z1)=2π3, then z is

A
12+i32
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B
3+i
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C
1+i3
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D
32+i12
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Solution

The correct option is A 12+i32
Let z=x+iy
Given, arg(z)=π3
yx=tan(argz)=3
y=x3

Now, z1=(x+iy)1=(x1)+iy
Since, arg(z1)=2π3, therefore z1 lies in 2nd quadrant in argand plane.
x1<0 and y>0
Now πα=arg(z1)
α=π3
tanα=yx1
3=yx1
y1x=3
x3=3(1x)
2x=1
x=12 and y=32

Thus, the complex number is 12+i32

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