We represent complex numbers in trigonometric form:
z 1 = r 1 ( cos φ + i sin φ ) , z 2 = r 2 ( cos ψ + i sin ψ ) {z_1} = {r_1}\left( {\cos \varphi + i\sin \varphi } \right),\,\,{z_2} = {r_2}\left( {\cos \psi + i\sin \psi } \right) z 1 = r 1 ( cos φ + i sin φ ) , z 2 = r 2 ( cos ψ + i sin ψ ) r 1 = ∣ z 1 ∣ , r 2 = ∣ z 2 ∣ {r_1} = |{z_1}|,\,\,{r_2} = |{z_2}| r 1 = ∣ z 1 ∣ , r 2 = ∣ z 2 ∣
Then
z 1 + z 2 = r 1 cos φ + r 2 cos ψ + i ( r 1 sin φ + r 2 sin ψ ) {z_1} + {z_2} = {r_1}\cos \varphi + {r_2}\cos \psi + i\left( {{r_1}\sin \varphi + {r_2}\sin \psi } \right) z 1 + z 2 = r 1 cos φ + r 2 cos ψ + i ( r 1 sin φ + r 2 sin ψ )
Then
∣ z 1 + z 2 ∣ 2 = ( r 1 cos φ + r 2 cos ψ ) 2 + ( r 1 sin φ + r 2 sin ψ ) 2 = r 1 2 cos 2 φ + 2 r 1 r 2 cos φ cos ψ + r 2 2 cos 2 ψ + r 1 2 sin 2 φ + 2 r 1 r 2 sin φ sin ψ + r 2 2 sin 2 ψ = r 1 2 ( cos 2 φ + sin 2 φ ) + r 2 2 ( cos 2 ψ + sin 2 ψ ) + 2 r 1 r 2 ( cos φ cos ψ + sin φ sin ψ ) = r 1 2 + r 2 2 + 2 r 1 r 2 ( cos φ cos ψ + sin φ sin ψ ) = r 1 2 + r 2 2 + 2 r 1 r 2 cos ( φ − ψ ) |{z_1} + {z_2}{|^2} = {\left( {{r_1}\cos \varphi + {r_2}\cos \psi } \right)^2} + {\left( {{r_1}\sin \varphi + {r_2}\sin \psi } \right)^2} = r_1^2{\cos ^2}\varphi + 2{r_1}{r_2}\cos \varphi \cos \psi + r_2^2{\cos ^2}\psi + r_1^2{\sin ^2}\varphi + 2{r_1}{r_2}\sin \varphi \sin \psi + r_2^2{\sin ^2}\psi = r_1^2\left( {{{\cos }^2}\varphi + {{\sin }^2}\varphi } \right) + r_2^2\left( {{{\cos }^2}\psi + {{\sin }^2}\psi } \right) + 2{r_1}{r_2}\left( {\cos \varphi \cos \psi + \sin \varphi \sin \psi } \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}\left( {\cos \varphi \cos \psi + \sin \varphi \sin \psi } \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {\varphi - \psi } \right) ∣ z 1 + z 2 ∣ 2 = ( r 1 cos φ + r 2 cos ψ ) 2 + ( r 1 sin φ + r 2 sin ψ ) 2 = r 1 2 cos 2 φ + 2 r 1 r 2 cos φ cos ψ + r 2 2 cos 2 ψ + r 1 2 sin 2 φ + 2 r 1 r 2 sin φ sin ψ + r 2 2 sin 2 ψ = r 1 2 ( cos 2 φ + sin 2 φ ) + r 2 2 ( cos 2 ψ + sin 2 ψ ) + 2 r 1 r 2 ( cos φ cos ψ + sin φ sin ψ ) = r 1 2 + r 2 2 + 2 r 1 r 2 ( cos φ cos ψ + sin φ sin ψ ) = r 1 2 + r 2 2 + 2 r 1 r 2 cos ( φ − ψ )
But
cos ( φ − ψ ) ≤ 1 \cos \left( {\varphi - \psi } \right) \le 1 cos ( φ − ψ ) ≤ 1
Then
r 1 2 + r 2 2 + 2 r 1 r 2 cos ( φ − ψ ) ≤ r 1 2 + r 2 2 + 2 r 1 r 2 ⇒ ∣ z 1 + z 2 ∣ 2 ≤ ( r 1 + r 2 ) 2 r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {\varphi - \psi } \right) \le r_1^2 + r_2^2 + 2{r_1}{r_2} \Rightarrow |{z_1} + {z_2}{|^2} \le {\left( {{r_1} + {r_2}} \right)^2} r 1 2 + r 2 2 + 2 r 1 r 2 cos ( φ − ψ ) ≤ r 1 2 + r 2 2 + 2 r 1 r 2 ⇒ ∣ z 1 + z 2 ∣ 2 ≤ ( r 1 + r 2 ) 2
Since
∣ z 1 + z 2 ∣ ≥ 0 , r 1 ≥ 0 , r 2 ≥ 0 |{z_1} + {z_2}| \ge 0,\,\,{r_1} \ge 0,\,\,{r_2} \ge 0 ∣ z 1 + z 2 ∣ ≥ 0 , r 1 ≥ 0 , r 2 ≥ 0
then
∣ z 1 + z 2 ∣ ≤ r 1 + r 2 ⇒ ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ |{z_1} + {z_2}| \le {r_1} + {r_2} \Rightarrow |{z_1} + {z_2}| \le |{z_1}| + |{z_2}| ∣ z 1 + z 2 ∣ ≤ r 1 + r 2 ⇒ ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣
Q. E. D
If
z 1 = 6 , z 2 = 4 + 3 i z_{1}=6,z_{2}=4+3i z 1 = 6 , z 2 = 4 + 3 i
then
∣ z 1 ∣ = 6 , ∣ z 2 ∣ = 4 2 + 3 2 = 5 |{z_1}| = 6,\,|{z_2}| = \sqrt {{4^2} + {3^2}} \, = 5 ∣ z 1 ∣ = 6 , ∣ z 2 ∣ = 4 2 + 3 2 = 5
but then
∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ = 6 + 5 = 11 ⇒ ∣ z 1 + z 2 ∣ ≤ 11 |{z_1} + {z_2}| \le |{z_1}| + |{z_2}| = 6 + 5 = 11 \Rightarrow |{z_1} + {z_2}| \le 11 ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ = 6 + 5 = 11 ⇒ ∣ z 1 + z 2 ∣ ≤ 11
Q. E. D
In general, the least value ∣ z 1 + z 2 ∣ = 0 \left| {{z_1} + {z_2}} \right| = 0 ∣ z 1 + z 2 ∣ = 0 z 1 = z 2 = 0 {z_1} = {z_2} = 0 z 1 = z 2 = 0
If z 1 = 6 , z 2 = 4 + 3 i z_{1}=6,z_{2}=4+3i z 1 = 6 , z 2 = 4 + 3 i
∣ z 1 + z 2 ∣ = ∣ 6 + 4 + 3 i ∣ = ∣ 10 + 3 i ∣ = 10 2 + 3 2 = 109 \left| {{z_1} + {z_2}} \right| = \left| {6 + 4 + 3i} \right| = \left| {10 + 3i} \right| = \sqrt {{{10}^2} + {3^2}} = \sqrt {109} ∣ z 1 + z 2 ∣ = ∣ 6 + 4 + 3 i ∣ = ∣ 10 + 3 i ∣ = 10 2 + 3 2 = 109
#339914 on Dec 2023
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