Question

Halla la mayor parte real posible de

(75 + 117\mathrm{i})z + \frac{96 + 144\mathrm{i}}{z}

donde

z

es un número complejo con

|z| = 4.

Aquí

\mathrm{i} = \sqrt{-1}.

Find the largest possible real part of

(75 + 117\mathrm{i})z + \frac{96 + 144\mathrm{i}}{z}

where

z

is a complex number with

|z| = 4.

Here

\mathrm{i} = \sqrt{-1}.

Respuesta

Escribe

z = 4(\cos\theta + \mathrm{i}\sin\theta)

y toma la parte real de cada término

Write

z = 4(\cos\theta + \mathrm{i}\sin\theta)

and take the real part of each term

Obtienes una expresión de la forma

a\cos\theta + b\sin\theta,

cuyo máximo es

\sqrt{a^2 + b^2}

You get an expression of the form

a\cos\theta + b\sin\theta,

whose maximum is

\sqrt{a^2 + b^2}

Escribe

z = 4(\cos\theta + \mathrm{i}\sin\theta),

así que

\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta).

La parte real de

(75 + 117\mathrm{i})z

es

4(75\cos\theta - 117\sin\theta)

= 300\cos\theta - 468\sin\theta,

y la parte real de

(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)

es

24\cos\theta + 36\sin\theta.

La parte real total es

324\cos\theta - 432\sin\theta,

cuyo máximo sobre

\theta

es

\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540. \end{aligned}

Write

z = 4(\cos\theta + \mathrm{i}\sin\theta),

so

\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta).

The real part of

(75 + 117\mathrm{i})z

is

4(75\cos\theta - 117\sin\theta)

= 300\cos\theta - 468\sin\theta,

and the real part of

(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)

is

24\cos\theta + 36\sin\theta.

The total real part is

324\cos\theta - 432\sin\theta,

whose maximum over

\theta

is

\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540. \end{aligned}