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Question: The amplitude of $\frac{(1+i)}{(1+i\sqrt{3})}$...

The amplitude of (1+i)(1+i3)\frac{(1+i)}{(1+i\sqrt{3})}

Answer

−π12-\frac{\pi}{12}

Explanation

Solution

To find the amplitude (argument) of the complex number Z=(1+i)(1+i3)Z = \frac{(1+i)}{(1+i\sqrt{3})}, we can use the property that arg⁡(Z1Z2)=arg⁡(Z1)−arg⁡(Z2)\arg\left(\frac{Z_1}{Z_2}\right) = \arg(Z_1) - \arg(Z_2).

Let Z1=1+iZ_1 = 1+i and Z2=1+i3Z_2 = 1+i\sqrt{3}.

  1. Find the amplitude of Z1=1+iZ_1 = 1+i:
    The complex number 1+i1+i lies in the first quadrant.
    The argument arg⁡(Z1)\arg(Z_1) is given by tan⁡−1(imaginary partreal part)\tan^{-1}\left(\frac{\text{imaginary part}}{\text{real part}}\right).
    arg⁡(Z1)=tan⁡−1(11)=tan⁡−1(1)=π4\arg(Z_1) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4}.

  2. Find the amplitude of Z2=1+i3Z_2 = 1+i\sqrt{3}:
    The complex number 1+i31+i\sqrt{3} also lies in the first quadrant.
    The argument arg⁡(Z2)\arg(Z_2) is given by tan⁡−1(imaginary partreal part)\tan^{-1}\left(\frac{\text{imaginary part}}{\text{real part}}\right).
    arg⁡(Z2)=tan⁡−1(31)=tan⁡−1(3)=π3\arg(Z_2) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.

  3. Calculate the amplitude of Z=Z1Z2Z = \frac{Z_1}{Z_2}:
    arg⁡(Z)=arg⁡(Z1)−arg⁡(Z2)\arg(Z) = \arg(Z_1) - \arg(Z_2)
    arg⁡(Z)=π4−π3\arg(Z) = \frac{\pi}{4} - \frac{\pi}{3}
    To subtract these fractions, find a common denominator, which is 12:
    arg⁡(Z)=3π12−4π12\arg(Z) = \frac{3\pi}{12} - \frac{4\pi}{12}
    arg⁡(Z)=−π12\arg(Z) = -\frac{\pi}{12}

The principal amplitude (argument) is typically given in the interval (−π,π](-\pi, \pi]. Our result −π12-\frac{\pi}{12} falls within this interval.

Alternatively, we can convert the complex numbers to polar form first:
Z1=1+i=12+12(cos⁡(π4)+isin⁡(π4))=2(cos⁡(π4)+isin⁡(π4))Z_1 = 1+i = \sqrt{1^2+1^2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})) = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))
Z2=1+i3=12+(3)2(cos⁡(π3)+isin⁡(π3))=2(cos⁡(π3)+isin⁡(π3))Z_2 = 1+i\sqrt{3} = \sqrt{1^2+(\sqrt{3})^2}(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))

Then, Z=Z1Z2=2(cos⁡(π4)+isin⁡(π4))2(cos⁡(π3)+isin⁡(π3))Z = \frac{Z_1}{Z_2} = \frac{\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))}{2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))}
Using the property r1(cos⁡θ1+isin⁡θ1)r2(cos⁡θ2+isin⁡θ2)=r1r2(cos⁡(θ1−θ2)+isin⁡(θ1−θ2))\frac{r_1(\cos\theta_1 + i\sin\theta_1)}{r_2(\cos\theta_2 + i\sin\theta_2)} = \frac{r_1}{r_2}(\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)):
Z=22(cos⁡(π4−π3)+isin⁡(π4−π3))Z = \frac{\sqrt{2}}{2}\left(\cos\left(\frac{\pi}{4} - \frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right)\right)
Z=12(cos⁡(−π12)+isin⁡(−π12))Z = \frac{1}{\sqrt{2}}\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)
From this polar form, the amplitude is clearly −π12-\frac{\pi}{12}.