What is module of complex number (cos(`pi` /5)+isin(`pi` /5))^10?



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aruv's profile pic

Posted on (Answer #1)


By De-Moiver's Theorem


`=cos(pi/2)+isin(pi/2)`               (i)


`Z=r(cos(theta)+isin(theta))`         (ii)

Then mod(Z)=r .

Compare (i) and (ii), we have mod(z)=1.

Thus modulus of z=1

sciencesolve's profile pic

Posted on (Answer #2)

You need to evaluate the absolute value of the given complex number, hence, you need to use the following formula, such that:nometric

`|z| = sqrt(x^2 + y^2)`

You need to convert the trigonometric form of the complex number into the algebraic form to identify the real part and imaginary part, hence, you need to use De Moivre's theorem, such that:

`z = (cos theta + isin theta)^n = cos (n theta) + i*sin (n theta)`

Reasoning by analogy, yields:

`z = (cos(pi/5) + i*sin(pi/5))^10 => z = (cos(10pi/5) + i*sin(10pi/5)`

`z = cos(2pi) + i*sin (2pi)`

Hence, evaluating the real part `x` and the imaginary part `y` yields:

`x = cos(2pi) = 1`

`y = sin(2pi) = 0`

`z = 1 + i*0 => z = 1`

Evaluating the absolute value, yields:

`|z| = sqrt(1 + 0) => |z| = 1`

Hence, evaluating the absolute value of the given complex number, under the given conditions, yields `|z| = 1.`


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