Question

The two complex numbers z₁ and z₂ are represented on an argand diagram. Show that

Iz₁ + z₂I ≤ Iz₁I +Iz₂I. If Iz₁I = 6 and Iz₁I = 4+3i, Show that the greatest value of Iz₁ + z₂I is 11 and find its least value.

Answer 1

We represent complex numbers in trigonometric form:

${z_1} = {r_1}\left( {\cos \varphi + i\sin \varphi } \right),\,\,{z_2} = {r_2}\left( {\cos \psi + i\sin \psi } \right)$ (where ${r_1} = |{z_1}|,\,\,{r_2} = |{z_2}|$ )

Then

${z_1} + {z_2} = {r_1}\cos \varphi + {r_2}\cos \psi + i\left( {{r_1}\sin \varphi + {r_2}\sin \psi } \right)$

Then

$|{z_1} + {z_2}{|^2} = {\left( {{r_1}\cos \varphi + {r_2}\cos \psi } \right)^2} + {\left( {{r_1}\sin \varphi + {r_2}\sin \psi } \right)^2} = r_1^2{\cos ^2}\varphi + 2{r_1}{r_2}\cos \varphi \cos \psi + r_2^2{\cos ^2}\psi + r_1^2{\sin ^2}\varphi + 2{r_1}{r_2}\sin \varphi \sin \psi + r_2^2{\sin ^2}\psi = r_1^2\left( {{{\cos }^2}\varphi + {{\sin }^2}\varphi } \right) + r_2^2\left( {{{\cos }^2}\psi + {{\sin }^2}\psi } \right) + 2{r_1}{r_2}\left( {\cos \varphi \cos \psi + \sin \varphi \sin \psi } \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}\left( {\cos \varphi \cos \psi + \sin \varphi \sin \psi } \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {\varphi - \psi } \right)$

But

$\cos \left( {\varphi - \psi } \right) \le 1$

Then

$r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {\varphi - \psi } \right) \le r_1^2 + r_2^2 + 2{r_1}{r_2} \Rightarrow |{z_1} + {z_2}{|^2} \le {\left( {{r_1} + {r_2}} \right)^2}$

Since

$|{z_1} + {z_2}| \ge 0,\,\,{r_1} \ge 0,\,\,{r_2} \ge 0$

then

$|{z_1} + {z_2}| \le {r_1} + {r_2} \Rightarrow |{z_1} + {z_2}| \le |{z_1}| + |{z_2}|$

Q. E. D

If

$z_{1}=6,z_{2}=4+3i$

then

$|{z_1}| = 6,\,|{z_2}| = \sqrt {{4^2} + {3^2}} \, = 5$

but then

$|{z_1} + {z_2}| \le |{z_1}| + |{z_2}| = 6 + 5 = 11 \Rightarrow |{z_1} + {z_2}| \le 11$

Q. E. D

In general, the least value $\left| {{z_1} + {z_2}} \right| = 0$ (if ${z_1} = {z_2} = 0$ ).

If $z_{1}=6,z_{2}=4+3i$ then the least value is

$\left| {{z_1} + {z_2}} \right| = \left| {6 + 4 + 3i} \right| = \left| {10 + 3i} \right| = \sqrt {{{10}^2} + {3^2}} = \sqrt {109}$