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Chapter 2: Problem 131

\([\\{1+\cos (\pi / 12)+i \sin (\pi / 12)\\} /\\{1+\cos (\pi / 12)\) \(-i \sin (\pi / 12)\\}]^{36}=\) (a) \(-1\) (b) 1 (c) 0 (d) \((1 / 2)\)

Short Answer

Expert verified
The given complex expression simplifies to \(1\), and when raised to the power of \(36\), it remains \(1\), which corresponds to option (b).

Step by step solution

01

Simplify the Expression

To start, let's simplify the given expression: \[ \frac{1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})}{1+\cos(\frac{\pi}{12})-i\sin(\frac{\pi}{12})} \] One way to rationalize this complex number is to multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})) (1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12}))}{(1+\cos(\frac{\pi}{12})-i\sin(\frac{\pi}{12}))(1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12}))} \]
02

Multiply and Simplify

Next, we multiply and simplify the expression: \[ \frac{1+2\cos(\frac{\pi}{12})+\cos^2(\frac{\pi}{12}) - \sin^2(\frac{\pi}{12})}{1+2\cos(\frac{\pi}{12})+\cos^2(\frac{\pi}{12})+\sin^2(\frac{\pi}{12})} \] Now, note that \(\cos^2(\frac{\pi}{12})+\sin^2(\frac{\pi}{12})=1\) . So, we can replace this sum with 1: \[ \frac{1+2\cos(\frac{\pi}{12})+(1- \sin^2(\frac{\pi}{12}))}{1+2\cos(\frac{\pi}{12})+1} \] Further simplifying, we obtain: \[ \frac{2(1+\cos(\frac{\pi}{12}))}{2+2\cos(\frac{\pi}{12})} \]
03

Cancel terms

Now, cancel the common terms in the numerator and the denominator: \[ \frac{1+\cos(\frac{\pi}{12})}{1+\cos(\frac{\pi}{12})} = 1 \]
04

Raise to the 36th power

Now, we need to raise the result we found in the previous step to the power of 36: \[ (1)^{36} = 1 \] The result matches option (b) which is 1.

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Most popular questions from this chapter

If \(\mathrm{w}\) is one of the cube root of 1 other then 1 then $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1-\mathrm{w}^{2} & \mathrm{w}^{2} \\ 1 & \mathrm{w}^{2} & \mathrm{w}^{4} \end{array}\right|= $$ (a) \(3 \mathrm{w}\) (b) \(3 \mathrm{w}(\mathrm{w}-1)\) (c) \(3 \mathrm{w}^{2}\) (d) \(3 \mathrm{w}(1-\mathrm{w})\)

If \(Z=[(4+3 i) /(5-3 i)]\) then \(Z^{-1}=\) (a) \((11 / 25)-(27 / 25) \mathrm{i}\) (b) \([(-11) / 25)]-(27 / 25) \mathrm{i}\) (c) \([(-11) / 25)]+(27 / 25) \mathrm{i}\) (d) \((11 / 25)+(27 / 25) \mathrm{i}\)

If \(z=\left[(1+7 i) /(2-i)^{2}\right]\) then the polar form of \(z\) is (a) \(\sqrt{2}[\cos (3 \pi / 4)+\mathrm{i} \sin (3 \pi / 4)]\) (b) \(\sqrt{2}[\cos (\pi / 4)+\mathrm{i} \sin (\pi / 4)]\) (c) \(\sqrt{2}[\cos (7 \pi / 4)+\mathrm{i} \sin (7 \pi / 4)]\) (d) \(\sqrt{2}[\cos (5 \pi / 4)+\mathrm{i} \sin (5 \pi / 4)]\)

If \(\mathrm{z}_{1}=2-\mathrm{i}\) and \(\mathrm{z}_{2}=1+\mathrm{i}\) then \(\left|\left[\left(z_{1}-z_{2}+1\right) /\left(z_{1}+z_{2}+i\right)\right]\right|=\) (a) \(\sqrt{(5 / 3)}\) (b) \(\sqrt{(3 / 5)}\) (c) \(\sqrt{(4 / 5)}\) (d) \(\sqrt{(5 / 4)}\)

If \(1, \mathrm{w}\) and \(\mathrm{w}^{2}\) are cube root of 1 then \((1-\mathrm{w})\left(1-\mathrm{w}^{2}\right)\) \(\left(1-\mathrm{w}^{4}\right)\left(1-\mathrm{w}^{8}\right)=\) (a) 16 (b) 8 (c) 9 (d) 64
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