Argument of the summ of two complex numbersLet r, s be positive real numbers and θ, ϕ real numbers with |θ−ϕ|&lt;π. Then an argument of reiθ+seiϕ lies between θ and ϕCan someone give a short, clean proof of the statement above that doesn't rely on geometric intuition?

Question

Argument of the summ of two complex numbersLet r, s be positive real numbers and θ, ϕ real numbers with |θ−ϕ|&amp;lt;π. Then an argument of reiθ+seiϕ lies between θ and ϕCan someone give a short, clean proof of the statement above that doesn&#039;t rely on geometric intuition?

nick1337 · Accepted Answer

Step 1 You want to show that if you write reiθ+seiθ as teiχ, then χ lies between θ and ϕ. Without loss of generality, assume θ≤ϕ. Factoring out eiθ you get reiθ+seiϕ=eiθ(r+sei(ϕ−θ)). Since multiplying by eiθ is just a rotation by an angle θ, it is enough to consider the case where θ=0 and 0&amp;lt;ϕ&amp;lt;π. In that case, you have r+seiϕ=r+s(cos⁡(ϕ)+isin⁡(ϕ)), and you want to express it in the form t(cos⁡(χ)+isin⁡(χ)) Looking at real and complex parts, you see that tsin⁡(χ)=ssin⁡(ϕ), and r+scos⁡(ϕ)=tcos⁡(χ) Assume first that 0&amp;lt;ϕ&amp;lt;π2. Then χ must also lie in the first quadrant, since we need both sin⁡(χ) and cos⁡(χ)to be positive (since t, ssin⁡(ϕ), and r+scos⁡(ϕ) are all positive). If, on the other hand, π2&amp;lt;ϕ&amp;lt;π, then cos⁡(ϕ) is negative. If r+scos⁡(ϕ)&amp;gt;0, then we need cos⁡(χ)&amp;gt;0, so χ is in the first quadrant and automatically smaller than ϕ and we are done. If r+scos⁡(ϕ) is negative then we need χ in the second quadrant. In the former case, 0&amp;lt;χ, ϕ&amp;lt;π2, then from sin⁡(χ)=stsin⁡(ϕ) and since sin⁡(x) is increasing on 0≤x≤π2, then 0&amp;lt;χ&amp;lt;ϕ if and only if st&amp;lt;1, if and only if s&amp;lt;t. Now note that t2=||r+seiϕ||2=r2+s2+2rscos⁡(ϕ)&amp;gt;s2, since all of r, s, and cos⁡(ϕ) are positive. Since s and t are both positive, s&amp;lt;t, which shows that 0&amp;lt;χ&amp;lt;ϕ, as desired. In the other case, where π2&amp;lt;ϕ&amp;lt;π and r+scos⁡(ϕ)&amp;lt;0, then we know χ is also in the second quadrant where sin⁡(x) is decreasing, so from sin⁡(χ)=stsin⁡(ϕ) we get that π2&amp;lt;χ&amp;lt;ϕ if and only if st&amp;gt;1, if and only if s&amp;gt;t. Here, since cos⁡(θ)&amp;lt;0, then you have again t2=||r+seiϕ||2=s2+r(r+2s cos⁡(ϕ))&amp;lt;s2 (since r+scos⁡(ϕ)&amp;lt;0 in this situation), so you get t&amp;lt;s and hence π2&amp;lt;χ&amp;lt;ϕ, as desired

Vasquez · Accepted Answer

It&#039;s enough to do the case θ=0 (just factor out eiθ). Assume 0&amp;lt;ϕ&amp;lt;π2. Then0&amp;lt;tan⁡arg⁡(r+seiϕ)=ssin⁡ϕr+scos⁡ϕ&amp;lt;ssin⁡ϕscos⁡ϕ=tan⁡ϕso the argument of the sum is between 0 and ϕ as desired. Something similar should work when ϕ is obtuse.

Argument of the summ of two complex numbers Let r, s

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