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F(z)=1+i+21​+i+…+1/n​−i, then the principal argument of z can be:

(a) π/2​
(b) π/4​
(c) π/6​
(d) π/3​

User Yotam Omer
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1 Answer

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The principal argument of z is 0, which corresponds to choice (a) π/2 .

The expression F(z)=1+i+ 2/ 1+i​ +⋯+ 1/n −i involves a sequence of complex numbers being added together.

Let's simplify the expression for F(z) step by step:

F(z)=1+i+ 2/ 1+i​ +⋯+ 1/n −i

To simplify 2/ 1+i​ , multiply the numerator and denominator by the conjugate of the denominator to eliminate the complex number from the denominator:

2/ 1+i = 2(1−i)/ (1+i)(1−i) = 2−2i/ 1+1 =1−i

So, the expression becomes:

F(z)=1+i+(1−i)+⋯+ 1/n −i

Simplifying further:

F(z)=2+ 1/n

Now, let's analyze the expression F(z):

F(z)=2+ 1/n

​As n approaches infinity, 1/n approaches zero, and F(z) approaches 2.

The complex number z represented by this expression must lie on the line passing through the origin and the point 2 in the complex plane. The argument of z can be found as the angle made by this line with the positive real axis.

The argument of z can be calculated as:

Argument of z=arctan( Re(z)/Im(z) )

For the point 2 on the complex plane, the real part is 2 and the imaginary part is 0. So, the argument of z is:

Argument of z=arctan( 0/2 )=arctan(0)=0

Therefore, the principal argument of z is 0, which corresponds to choice (a) π/2 .

User Seanlinsley
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