# Complex Number - Finding the Modulus

• Nov 8th 2010, 06:47 PM
Paymemoney
Complex Number - Finding the Modulus
Hi

Can someone tell me what is wrong with my answer to the following question

Find the modulus: $j(1+jcos\theta+sin\theta)$

$=j+j^2cos\theta+jsin\theta$
$=j-cos\theta+jsin\theta$
$=\sqrt{j^2+cos\theta^2+j^2sin\theta^2}$
$=\sqrt{-1+cos\theta^2-sin\theta^2}$
$=\sqrt{-1+1-sin\theta^2-sin\theta^2}$
$=\sqrt{-2sin\theta^2}$

P.S
• Nov 8th 2010, 06:51 PM
Prove It
First, write your complex number in terms of the real and imaginary parts:

$\displaystyle j(1 + j\cos{\theta} + \sin{\theta}) = j + j^2\cos{\theta} + j\sin{\theta}$

$\displaystyle = j - \cos{\theta} + j\sin{\theta}$

$\displaystyle = -\cos{\theta} + j(1 + \sin{\theta})$.

So the modulus is

$\displaystyle \sqrt{(-\cos{\theta})^2 + (1 + \sin{\theta})^2}$

$\displaystyle = \sqrt{\cos^2{\theta} + 1 + 2\sin{\theta} + \sin^2{\theta}}$

$\displaystyle = \sqrt{1 + 1 + 2\sin{\theta}}$

$\displaystyle = \sqrt{2 + 2\sin{\theta}}$

$\displaystyle = \sqrt{2(1 + \sin{\theta})}$.