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Exercise 2.7 · Q3

Q.If (x1+iy1)(x2+iy2)(x3+iy3)⋯(xn+iyn)=a+ib(x_1+iy_1)(x_2+iy_2)(x_3+iy_3)\cdots(x_n+iy_n)=a+ib, show that

(i) (x12+y12)(x22+y22)(x32+y32)⋯(xn2+yn2)=a2+b2(x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)\cdots(x_n^2+y_n^2)=a^2+b^2
(ii) ∑r=1ntan⁡−1(yrxr)=tan⁡−1(ba)+2kπ, k∈Z\displaystyle\sum_{r=1}^n\tan^{-1}\left(\dfrac{y_r}{x_r}\right)=\tan^{-1}\left(\dfrac ba\right)+2k\pi,\ k\in\mathbb Z.
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Starting from the given product identity (x1+iy1)(x2+iy2)⋯(xn+iyn)=a+ib(x_1+iy_1)(x_2+iy_2)\cdots(x_n+iy_n)=a+ib, we apply the modulus property ∣z1z2⋯zn∣=∣z1∣∣z2∣⋯∣zn∣|z_1z_2\cdots z_n|=|z_1||z_2|\cdots|z_n| for part (i), and the argument property arg⁡(z1z2⋯zn)=arg⁡z1+arg⁡z2+⋯+arg⁡zn\arg(z_1z_2\cdots z_n)=\arg z_1+\arg z_2+\cdots+\arg z_n for part (ii).

Step 1. (i) Take the modulus of both sides. For each factor, ∣xr+iyr∣=xr2+yr2|x_r+iy_r|=\sqrt{x_r^2+y_r^2}, and by the modulus-of-a-product property (§2.5.1, Property 3, extended to nn factors),

∣(x1+iy1)(x2+iy2)⋯(xn+iyn)∣=∣x1+iy1∣∣x2+iy2∣⋯∣xn+iyn∣.|(x_1+iy_1)(x_2+iy_2)\cdots(x_n+iy_n)|=|x_1+iy_1||x_2+iy_2|\cdots|x_n+iy_n|.

So

x12+y12⋅x22+y22⋯xn2+yn2=∣a+ib∣=a2+b2.\sqrt{x_1^2+y_1^2}\cdot\sqrt{x_2^2+y_2^2}\cdots\sqrt{x_n^2+y_n^2}=|a+ib|=\sqrt{a^2+b^2}.

Step 2. (i) Square both sides. Squaring removes all the square roots at once:

(x12+y12)(x22+y22)⋯(xn2+yn2)=a2+b2.(x_1^2+y_1^2)(x_2^2+y_2^2)\cdots(x_n^2+y_n^2)=a^2+b^2.

This proves part (i).

Step 3. (ii) Take the argument of both sides. By the argument-of-a-product property (arg z1z2=z_1z_2= arg z1+z_1+ arg z2z_2, extended to nn factors and understood as a general argument, i.e. equal only up to an integer multiple of 2π2\pi):

arg⁡[(x1+iy1)(x2+iy2)⋯(xn+iyn)]=arg⁡(x1+iy1)+arg⁡(x2+iy2)+⋯+arg⁡(xn+iyn).\arg\big[(x_1+iy_1)(x_2+iy_2)\cdots(x_n+iy_n)\big]=\arg(x_1+iy_1)+\arg(x_2+iy_2)+\cdots+\arg(x_n+iy_n). …

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