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If $3 - 5i$ is a square root of $z$, find the other root.

Well, I was under the impression that only the sign in front of the imaginary part would change so the other root would be $3 + 5i$.

However, when I solve for the complex root using de Moivre's theorem, then I get $-3 + 5i$.

Both the real and imaginary parts have changed signs - this seems to go against what I thought was the complex conjugate root theorem?

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  • $\begingroup$ would you like to show your working so someone can see what went wrong? $\endgroup$ – Siong Thye Goh Feb 4 at 4:03
  • $\begingroup$ Consider: is $z$ real or complex? $\endgroup$ – J. W. Tanner Feb 4 at 4:06
  • $\begingroup$ Z is complex - although it does not say, this is the assumption as the unit is on Demoivres. Aah so does that make a difference? If this is the case, is it not possible to use Demoivre's to solve for all roots if z is real? $\endgroup$ – McMath Feb 4 at 4:07
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    $\begingroup$ If $w$ is a square root, so is $-w$. So you have to negate $3$ and $5$. $\endgroup$ – D_S Feb 4 at 4:53
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We have the general property that: $t^2 =(-t)^2$. This holds in $\Bbb C$ as much as it does in $\Bbb R$. Hence if $z=(3-5i)^2$ then $z=(-(3-5i))^2=(-3+5i)^2$ holds also. On the other hand, we have:

$$(\bar{z})^2=\overline {z^2},$$

which is obvious if you square out $a+bi$ and $a-bi$ or represent them and their squares on an Argand Diagram.

The complex conjugate root theorem is based on

$$(x-z)(x-\bar z)=x^2-(z+\bar z)x+ z\bar z ,$$

for which all coefficients are real, thusly we can see complex solutions to real polynomials. However, the polynomial we might generate from this problem;

$$z^2+16+30i=0$$ does not have real coefficients, so that method can't be used.

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If we apply DeMoivre's Theorem, the second root will be a $180$ degree rotation around the pole. So the other root is in fact, $-3 + 5i$ (you're right!). You are confusing the usage of the complex conjugate root theorem, which only applies to polynomials with real coefficients, not square roots of complex numbers in the complex plane.

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  • $\begingroup$ I see, so it is not possible to use DeMoivre's theorem for polynomials with real coefficients? What if I was asked to find all the cube roots of 8 - how would I know if I am able to use this theorem? $\endgroup$ – McMath Feb 4 at 4:13
  • $\begingroup$ @McMath It's possible in some scenarios to use DeMoivre's Theorem for certain polynomials (like when solving for $x^n = a$ for some $a \in \mathbb{R}$), but more generally speaking, it finds and guarantees $k$ solutions to the complex equation $z^k = b$ for any $b \in \mathbb{C}$. $\endgroup$ – Hyperion Feb 4 at 4:17
  • $\begingroup$ @McMath you are most certainly right to use DeMoivre's Theorem to find the cube of $8$, but I'm just saying that you are confusing the use of DeMoivre's Theorem with the complex conjugate root theorem. $\endgroup$ – Hyperion Feb 4 at 4:20
  • $\begingroup$ Ok, so for finding all the cube roots of 8, I would be able to use DeMoivres and it would generate the same roots regardless of if I'm in the complex plane or real plane? Also, if I'm given a complex root like the root given in my original post, then I'm assuming the teacher would need to specify that z is complex, otherwise I could use the complex conjugate theorem to generate different roots. Thanks $\endgroup$ – McMath Feb 4 at 4:22
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    $\begingroup$ Cube roots of $8$: You want to find all solutions of $X^3=8$, in other words the solutions of $X^3-8=0$. Just factor, to get $(X-2)(X^2+2X+4)$. Use quadratic formula to find the roots of the quadratic factor. Please do this computation, and see that your three answers are the same as what you get using DeMoivre. $\endgroup$ – Lubin Feb 4 at 4:57
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Polar coordinates:

$z_1= re^{i\alpha}$, given.

$z_1^2= r^2e^{i(2\alpha)}=: z$.

The roots of $z$ are:

$z_1= re^{i(2\alpha)/2}$,and

$z_2= re^{i[(2\alpha)/2+ (2π)/2]}$;

$z_2= z_1\cdot(e^{iπ})= -z_1$.

Hence $z_2=-3+i5$.

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