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$z\bar{z}+3iz=p+9i$, where $z$ is a complex number and p is a real constant. Given this equation has exactly one root, determine the complex number $z$.

I started with $z=a+bi$ :

$$\begin{align}(a+bi)(a-bi)+3i(a+bi)&=p+9i\\a^2+b^2+3ai-3b&=p+9i\end{align}$$

Equating coefficients and solving for $a$ :

$$\begin{align}3a&=9\\a^2+b^2-3b&=p\\\implies b^2-3b+9&=p\end{align}$$

How can I continue this?

AnthonyML
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    This is just an ordinary quadratic equation in $b$, no? You also know that $3^2-4(1)(9-p)=0$ since there can only be one root. – Luke Collins Apr 08 '23 at 11:18
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    Write as $b^2-3b +(9-p)=0$ and use the discriminant $= 0$ condition for there to be exactly one root. – mcd Apr 08 '23 at 11:19

4 Answers4

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What you did so far is perfect. Now since you also know there is only one root, the equation for $b$ can have only one solution. This means the quadratic discriminant has to be zero: $$3^2-4(1)(9-p)=0\implies p = \frac{27}4.$$

Thus your equation is actually \begin{align*} b^2-3b+9&=\frac{27}4\implies(2b-3)^2=0\implies b=\frac32, \end{align*} so $z=3+\frac32i$.

Luke Collins
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4

Another way

$$z\bar{z}+3iz=p+9i \iff \bar{z}z-3i\bar z=p-9i$$

and subtracting/summing we obtain

  • $6i\Re(z)=18i \implies \Re(z)=3$

  • $2|z|^2 -6\Im(z)=2p \implies \Im(z)^2-3\Im(z)+9-p=0$

with the condition

$$9-4\left(9-p\right)=0 \implies p=\frac{27}4$$

$$\Im(z)^2-3\Im(z)+\frac 9 4=0 \iff \left(\Im(z)-\frac 3 2\right)^2=0$$

therefore

$$z=3+i\frac 32$$

user
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You can solve $$b^2-3b+9-p=0$$ like any other quadratic equation:

$$b_{1,2} = \frac{3\pm\sqrt{9-4(9-p)}}{2} = \frac{3\pm\sqrt{4p-27}}{2}$$

Nuke_Gunray
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Yet another way, $z=0$ is not a solution, and multiplying by $\bar z \ne 0$ gives:

$$|z|^2 \,\bar z + 3 i \,|z|^2 \,=\, (p + 9i) \,\bar z \quad\iff\quad 3 i \,|z|^2 = (p + 9i - |z|^2) \,\bar z$$

Taking complex conjugates on both sides:

$$- 3 i |z|^2 = (p - 9i - |z|^2) \, z$$

Multiplying with $\,3 i \,z \,=\, p + 9i - |z|^2\,$ from the original equation and canceling $\,z\ne0\,$:

$$ \require{cancel} 9\,\cancel{z}\,|z|^2 = (p - 9i - |z|^2)(p + 9i - |z|^2)\,\cancel{z} \quad\iff\quad |z|^4 - (2p + 9)\,|z|^2 + p^2+ 81 = 0 $$

For a unique solution the discriminant of the quadratic must be zero:

$$ 0 = (2p+9)^2 - 4(p^2+81) = \cancel{4p^2}+36p+81-\cancel{4p^2}-324= 9(4p-27) \;\implies\; p = \frac{27}{4} $$

It follows that $\,|z|^2 = \dfrac{2p+9}{2}\Big|_{\,p\,=\,27/4} = \dfrac{45}{4}\,$, then from the original equation:

$$ 3 i \,z \,=\, p + 9i - |z|^2 = \frac{27}{4} + 9 i - \frac{45}{4}=\frac{9}{2}(-1+2i) \quad\implies\quad z = \frac{3}{2}(2+i) $$

dxiv
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