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Exercise 2.7 · Q4

Q.If 1+z1−z=cos⁡2θ+isin⁡2θ\dfrac{1+z}{1-z}=\cos2\theta+i\sin2\theta, show that z=itan⁡θz=i\tan\theta.

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We treat cos⁡2θ+isin⁡2θ\cos2\theta+i\sin2\theta as a single quantity ww, solve the given linear relation for zz algebraically, then simplify w−1w-1 and w+1w+1 each as a product with a common factor cos⁡θ+isin⁡θ\cos\theta+i\sin\theta that cancels.

Step 1. Let w=cos⁡2θ+isin⁡2θw=\cos2\theta+i\sin2\theta and solve 1+z1−z=w\dfrac{1+z}{1-z}=w for zz.

1+z=w(1−z)=w−wz ⇒ z+wz=w−1 ⇒ z(1+w)=w−1 ⇒ z=w−1w+1.1+z=w(1-z)=w-wz \ \Rightarrow\ z+wz=w-1 \ \Rightarrow\ z(1+w)=w-1 \ \Rightarrow\ z=\frac{w-1}{w+1}.

Step 2. Simplify w−1w-1 using cos⁡2θ−1=−2sin⁡2θ\cos2\theta-1=-2\sin^2\theta and sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta.

w−1=(cos⁡2θ−1)+isin⁡2θ=−2sin⁡2θ+2isin⁡θcos⁡θ=2sin⁡θ(−sin⁡θ+icos⁡θ).w-1=(\cos2\theta-1)+i\sin2\theta=-2\sin^2\theta+2i\sin\theta\cos\theta=2\sin\theta(-\sin\theta+i\cos\theta).

Since i(cos⁡θ+isin⁡θ)=icos⁡θ−sin⁡θ=−sin⁡θ+icos⁡θi(\cos\theta+i\sin\theta)=i\cos\theta-\sin\theta=-\sin\theta+i\cos\theta, this becomes

w−1=2sin⁡θ⋅i(cos⁡θ+isin⁡θ).w-1=2\sin\theta\cdot i(\cos\theta+i\sin\theta).

Step 3. Simplify w+1w+1 using cos⁡2θ+1=2cos⁡2θ\cos2\theta+1=2\cos^2\theta. …

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