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Solve for $z$, and give your answer in the form $a+bi$.

$$\overline {(z+1+5i)}= 3z+9−9i$$

I have gotten to where I have:

$$z=3\bar z+8+4i$$

But I am stuck here and could use some guidance.

Toby Mak
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LTD429
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    Use \overbar to put the entire left side under the bar; if you just want the bar over the z, use \bar, i.e. $\bar z$. – bjcolby15 Mar 29 '22 at 23:44
  • "I have gotten to where I have: $;z=3\bar z+8+4i$" $;\dots;$then write it as $,z-3\bar z = 8+4i,$, or $ 2,\underbrace{(z-\bar z)}{=,2i, \text{Im}(z)} - \underbrace{(z+\bar z)}{=,2,\text{Re}(z)} = 8 + 4i$ $\implies\begin{cases}\begin{align}-2,\text{Re}(z) &= 8 \ 4i,\text{Im}(z) &= 4i\end{align}\end{cases} $ – dxiv Mar 30 '22 at 06:39

4 Answers4

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Let $z=a+ib$ with $a$ and $b$ real. Then you get $a+ib=3(a-ib)+8+4i$. Compare real and imaginary parts and you will see that $a=-4, b=1$. So the answer is $z=-4+i$.

Kavi Rama Murthy
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4

Given

$z=3\overline z+8+4i,$

next take the conjugate equation:

$\overline z=3z+8-4i.$

Eliminate $\overline z$ between these equations to get

$z=3(3z+8-4i)+8+4i$

and solve by ordinary algebraic methods.

$-8z=32-8i, z=-4+i.$

Oscar Lanzi
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  • A variation on this: since $ \ z + \overline{z} \ = \ 2·\mathfrak{Re}(z) \ $ and $ \ z - \overline{z} \ = \ 2·\mathfrak{Im}(z) \ \ , \ $ adding the two equations together gives us $ \ 2·\mathfrak{Re}(z) \ = \ 3·2·\mathfrak{Re}(z) \ + \ 16 \ $ and subtracting the second from the first yields $ \ 2·\mathfrak{Im}(z) \ = \ 3·(-2)·\mathfrak{Im}(z) \ + \ 8i \ \ . \ $ Hence, $$ 4·\mathfrak{Re}(z) \ + \ 16 \ \ = \ \ 0 \ \ \Rightarrow \ \ \mathfrak{Re}(z) \ \ = \ \ -4 $$ and $$ 8·\mathfrak{Im}(z) \ \ = \ \ 8i \ \ \Rightarrow \ \ \mathfrak{Im}(z) \ \ = \ \ i \ \ . $$ –  Jul 12 '23 at 08:04
2

We wish to solve the equation:

$$\overline {(z+1+5i)}= 3z+9−9i$$

(Here, $z \in \mathbf{C}$)

We simplify the given equation as $$ \bar{z} + 1 - 5 i = 3 z + 9 - 9 i $$ i.e. $$ \bar{z} - 3 z + 4 i - 8 = 0 \tag{1} $$

Let $z = x + i y$. Then $\bar{z} = x - i y$.

Substituting above into (1), we get $$ x - i y - 3 (x + i y) + 4 i - 8 = 0 $$ or $$ - 2 x - 4 i y + 4 i - 8 = 0 $$ or $$ x + 2 i y - 2 i + 4 = 0 $$ or $$ (x + 4) + 2 i (y - 1) = 0 \tag{2} $$

Equating the real and imaginary parts on both sides of (2), we get $$ x = -4 \ \ \mbox{and} \ \ y = 1 $$

Thus, the solution of the given equation is: $$ z = x + i y = -4 + i $$ i.e. $$ \boxed{z = -4 + i} $$

Dr. Sundar
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0

Not suggesting this as a quick way to solve the problem, just another way to interpret the equation:

Re-arranging the given equation as $$ \overline {z \ + \ 1 \ + \ 5i} \ \ = \ \ 3z \ + \ 9 \ − \ 9i \ \ \Rightarrow \ \ \overline{z} \ + \ \overline{1 \ + \ 5i} \ \ = \ \ 3z \ + \ 9 \ − \ 9i $$ $$ \Rightarrow \ \ \overline{z} \ + \ 1 \ - \ 5i \ \ = \ \ 3z \ + \ 9 \ − \ 9i \ \ \Rightarrow \ \ \overline{z} \ \ = \ \ 3z \ + \ 8 \ − \ 4i \ \ \ , $$

we can interpret this as saying that if the vector $ \ \vec{u} \ = \ \langle \ a \ , \ b \ \rangle \ $ represents $ \ z \ = \ a + bi \ $ in the Argand plane, then multiplying $ \ \vec{u} \ $ by $ \ 3 \ $ and adding $ \ \vec{v} \ = \ \langle \ 8 \ , \ -4 \ \rangle \ $ to it has the effect of changing the sign of the imaginary part; thus, $ \ 3·\vec{u} \ + \ \vec{v} \ = \ \langle \ a \ , \ -b \ \rangle \ \ . $

This means that the length of the vector (the modulus $ \ |z| \ ) \ $ is unchanged by these operations, or $ \ \Vert \ 3·\vec{u} \ + \ \vec{v} \ \Vert \ = \ \Vert \ \vec{u} \ \Vert \ \ . \ $ Since the addition of $ \ \vec{v} \ $ decreases the imaginary component of $ \ 3·\vec{u} \ , \ $ we can conclude that $ \ b \ > \ 0 \ \ ; \ $ similarly, since the real component is increased by this addition, we must have $ \ a \ < \ 0 \ $ (that is, $ \ z \ $ lies in the second quadrant and $ \ \overline{z} \ $ in the third).

The argument of $ \ z \ $ is changed from $ \ \theta \ $ to $ \ 2 \pi - \theta \ \ , \ $ so we have $ \ \cos (2 \pi - \theta) \ = \ \cos \theta \ \Rightarrow \ a \ = \ 3a + 8 \ $ and $ \ \sin (2 \pi - \theta) \ = \ -\sin \theta \ \Rightarrow \ -b \ = \ 3b - 4 \ \ , \ $ which equations lead to the same results as in other answers.

We can also apply the geometric interpretation above to write the change of sign of the tangent of the argument of $ \ z \ $ as $$ \frac{3b \ - \ 4}{3a \ + \ 8} \ \ = \ \ \frac{-b}{a} \ \ \Rightarrow \ \ 3ab \ - \ 4a \ \ = \ \ -3ab \ - \ 8b \ \ \Rightarrow \ \ 6ab \ \ = \ \ 4a \ - \ 8b $$ $$ \Rightarrow \ \ 1 \ \ = \ \ \frac{2}{3b} \ - \ \frac{4}{3a} \ \ . $$

The unchanged modulus-squared of $ \ z \ $ is expressed by $ \ (3a \ + \ 8)^2 \ + \ (3b \ - \ 4)^2 \ \ = \ \ a^2 \ + \ b^2 $ $ \Rightarrow \ \ 8a^2 + 48a + 8b^2 - 24b + 80 = 0 \ \ . \ $ Solving these two equations simultaneously (with a bit of labor) yields two solutions (in the second quadrant as expected) :

$$ \langle \ -4 \ , \ 1 \ \rangle \ \rightarrow \ z \ = \ -4 \ + \ i \ \ \ , \ \ \ \langle \ -2 \ , \ 2 \ \rangle \ \rightarrow \ z \ = \ -2 \ + \ 2i \ \ \ . $$

We find $ \ 3·\langle \ -4 \ , \ 1 \ \rangle + \langle \ 8 \ , \ 4 \ \rangle \ = \ \langle \ -12 + 8 \ , \ 3 - 4 \ \rangle \ = \ \langle \ -4 \ , \ -1 \ \rangle \ \ , \ $ giving $ \ \overline{z} \ $ as desired. For $ \ 3·\langle \ -2 \ , \ 2 \ \rangle + \langle \ 8 \ , \ 4 \ \rangle \ = \ \langle \ -6 + 8 \ , \ 6 - 4 \ \rangle \ = \ \langle \ 2 \ , \ 2 \ \rangle \ \ , \ $ the modulus is unaltered and the tangent of the argument has its sign changed, but the resulting vector does not corresponds to $ \ \overline{z} \ \ . \ $ So this second solution is spurious. [This is certainly not a fast method, but is included here to illustrate the "geometry" of the equation.]