Solving $z+i\overline{z}=iz-\overline{z}$

Question

I want to solve $z+i\overline{z}=iz-\overline{z}$ ($\overline{z}$ is the complex conjugate). I have solved it setting $z=a+bi$. But can one solve without writing it $z$ a certain form, factorization maybe?

Thanks in advance

$\frac{z+\bar z}{z-\bar z}=$purely imaginary. – Someone May 21 '15 at 13:48 — Someone, May 21 '15 at 13:48

score 2 · Answer 1 · answered May 21 '15 at 14:25

Let's rewrite the equation: $$ z+i\bar z=iz-\bar z \Longleftrightarrow \bar z(i+1)=z(i-1)\Longleftrightarrow z=\frac{i+1}{i-1}\bar z=-i\bar z\;\;\;. $$

Writing now $z=re^{i\theta}$, and observing that $i=e^{i\frac{\pi}2}$ the last equation becomes $$ re^{i\theta}=e^{i\frac{\pi}2}re^{-i\theta}\;\;\; $$ i.e. $$ e^{i\theta}=e^{i(\frac{\pi}2-\theta)}\;\;. $$ So the solutions are $z=re^{i\theta}$, for $r\ge0$ and $\theta\in\Bbb R$ such that $\theta=\frac{\pi}2-\theta+2k\pi$, i.e. $\theta=\frac{\pi}4+k\pi$.

So you have infinite solutions in modulus, but your argument must be $\frac{\pi}4$ or $\frac{5\pi}4$.

Thus the solutions are $z_r=r(1+i)$, $r\in\Bbb R$.

score 1 · Answer 2 · answered May 21 '15 at 13:52

1

Hint For $w = z + i \bar{z}$ we have $$\bar{w} = \overline{z + i\bar{z}} = \bar{z} - i z = -(iz - \bar{z}),$$ in which case we can rewrite the given equation as $$w = -\bar{w},$$ which is satisfied iff $w$ is purely imaginary.

answered May 21 '15 at 13:52

Travis Willse

99,363

What if $w=0$ ? – Prasun Biswas May 21 '15 at 13:56
@PrasunBiswas $0$ is purely imaginary, as it can be written $0 i$. – Travis Willse May 21 '15 at 13:57
I used to think that $0$ is purely real. Maybe it's considered as both purely real and purely imaginary? – Prasun Biswas May 21 '15 at 13:57
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Yes, it's real, as $\Im 0 = 0$, but it's also purely imaginary, as $\Re 0 = 0$. Of course, it's the only complex number with both properties. – Travis Willse May 21 '15 at 13:58
Shouldn't the condition simply be $z=-i \bar z$ $\iff$ $a=-b$ – Someone May 21 '15 at 13:59
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@Mann Yes, that's another perfectly good characterization (thought the r.h.s. makes explicit reference to the decomposition of $z$ into real and imaginary parts, which OP stated he'd like to avoid). – Travis Willse May 21 '15 at 14:10

score 1 · Answer 3 · answered May 21 '15 at 14:08

Alternatively (the benefit of hindsight helps here), one can define $\zeta$ by setting $$z = \zeta e^{-\pi i / 4},$$ in which case substituting in the original equation and rearranging gives $$2 \sqrt{2} \Im \zeta = 0.$$ This is satisfied iff $\zeta$ is real, so the solution set is $$\{t e^{-\pi i / 4} : t \in \Bbb R\}.$$ This is the line produced by rotating the real line about the origin $\frac{\pi}{4}$ radians clockwise, namely, the line through the origin that bisects the second and fourth quadrants.

Martigan · Answer 4 · 2015-05-21T14:48:00.557

0

You could write it $z(1-i)=i\bar z(i-1)$...

EDIT: since there is a question about where does it come from... It is coming from: $z+i\bar z=iz -\bar z$ that is $z-iz=-i\bar z -\bar z$, then $z(1-i)=-i(\bar z-i\bar z)=-i\bar z(1-i)$

edited May 21 '15 at 14:48

answered May 21 '15 at 13:56

Martigan

5,844

score 0 · Answer 5 · answered May 21 '15 at 13:57

0

From your given condition it can be inferred that,

$$\frac{z+\bar z}{z-\bar z}=i$$

Using componendo and dividendo,

$$\frac{z}{\bar z}=\frac{1+i}{i-1}$$

$$z=-i\bar z$$

answered May 21 '15 at 13:57

Someone

2,865

score 0 · Answer 6 · answered May 21 '15 at 14:37

Another answer that uses an important and fundamental property:

Note, $$z+\overline{z} = 2\operatorname{Re}z,$$ and $$z-\overline{z} = 2i\operatorname{Im}z.$$

Using this, we get

$$z+i\overline{z} = iz-\overline{z} \implies z+\overline{z} = i(z-\overline{z}).$$ This becomes $$2\operatorname{Re}z = i(2i\operatorname{Im}z)\\ \operatorname{Re}z = -\operatorname{Im}z.$$

Therefore, solutions must lie along the line $y=-x$ in the complex plane.

Let's check using rectangular form. $z = a+bi$.

$$a+bi+i(a-bi) = i(a+bi)-a+bi.$$ Let $b = -a$. $$a-ai+i(a+ai) = i(a-ai)-a-ai \\ a-ai+ai-a = ai+a-a-ai \\ 0 = 0.$$

score 0 · Answer 7 · answered May 21 '15 at 16:43

$$z+i\overline{z}=iz-\overline{z}\Longrightarrow$$

$$(a+bi)+(a-bi)i=(a+bi)i-(a-bi)\Longleftrightarrow$$ $$(a+b)(1+i)=(a+b)(-1+i)\Longleftrightarrow$$ $$a+b+ai+bi=-a-b+ai+bi\Longleftrightarrow$$ $$2a+2b=0\Longleftrightarrow$$ $$2(a+b)=0\Longleftrightarrow$$ $$a+b=0\Longleftrightarrow$$ $$\{_{b=-a}^{a=-b}$$

Solving $z+i\overline{z}=iz-\overline{z}$

7 Answers7