Complex variables - find an additional solution

**Ragnarok** · January 13th 2013, 12:03 PM

This is not really a calculus question but I am taking a course in basic complex analysis and this is from the first chapter.

Show that the indicated numbers satisfy the given equation and find an additional solution, $z_2$ .

$z^2+i=0, z_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

Obviously you can plug $z_1$ in and get a true statement, but how do you go about finding an additional solution?

**SworD** · January 13th 2013, 12:05 PM

Do you recall synthetic or polynomial division? One way is to factor out $z - z_1$ from $z^2 + i$ and you will get a linear equation.

Edit: you don't even need to go this complicated: just use the fact that $z^2 + i$ is an even function, if z1 is a root, so is -z1.

**farzero** · January 13th 2013, 12:06 PM

Use the fact that complex zeros come in conjugate pairs. just conjugate z1 and call that z2, and plug it in and it also is a zero.

**SworD** · January 13th 2013, 12:07 PM

No, the above won't work because the polynomial does not have real coefficients.

**farzero** · January 13th 2013, 12:11 PM

Why won't the conjugate of z1 work? It is indeed a zero when I do plugged it in. Perhaps there is some theory I'm not seeing here, all the op's question asks is to find an additional solution. And so z1 conjugate is that solution.

**SworD** · January 13th 2013, 12:14 PM

No it isn't. Recheck your calculations.

The theory your not seeing here is that roots won't magically always be conjugate pairs. I could just create an arbitrary polynomial which clearly violates that rule, for example, by multiplying out the following:

$(z + i)(z + 2i)$

You are missing the fact that the coefficients of the polynomial have to be real in order for the roots to come in conjugate pairs.

The other root in this case is actually the negative of z1, not the conjugate.

**farzero** · January 13th 2013, 12:32 PM

We could probably just multiply z1 by e^(2kpi) for k in Z? Thanks sword for fixing my error, I didn't multiply correctly lol. Do you see an error with this new idea, using the fact that e^(2kpi)=1 is an identity.

**Deveno** · January 13th 2013, 01:36 PM

Originally Posted by farzero

We could probably just multiply z1 by e^(2kpi) for k in Z? Thanks sword for fixing my error, I didn't multiply correctly lol. Do you see an error with this new idea, using the fact that e^(2kpi)=1 is an identity.

if it is an identity, this won't give you an additional answer.

look suppose $z_1$ is a root (any root) of $z^2 + c = 0$ .

this means $z_1^2 + c = 0$ .

therefore: $(-z_1)^2 + c = (-z_1)(-z_1) + c =$

$(-1)(z_1)(-1)(z_1) + c = (-1)(-1)z_1^2 + c = z_1^2 + c = 0$ , that is: $-z_1$ is ALSO a root of $z^2 + c = 0$ .

note that we discovered this without any information about "c".

**HallsofIvy** · January 14th 2013, 09:47 AM

Originally Posted by farzero

Why won't the conjugate of z1 work? It is indeed a zero when I do plugged it in. Perhaps there is some theory I'm not seeing here, all the op's question asks is to find an additional solution. And so z1 conjugate is that solution.

Again, you are completely wrong. $z_1= -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i$ . Its complex conjugate is $-\frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i$ . The square is $\frac{2}{4}- 2\frac{2}{4}i- \frac{2}{4}= i$ so that if z is that complex conjugate, $z^2+ i= 2i$ , not 0.

**HallsofIvy** · January 14th 2013, 09:56 AM

Yes, having already determined one root, the simplest way to find the other is "division". If you wished to solve $z^2+i= 0$ from the beginning, use polar form. This equation is the same as $z^2= -i$ and -i has modulus 1 and argument $3\pi/2$ (or $-\pi/4$ ). To find the square root, take the square root of the modulus (1) and divide the argument by 2: $(3\pi/2)/2= 3\pi/4$ so that one square root of -i is $cos(3\pi/4)+ i sin(3\pi/4)= -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i$ which is the root we have.

Since adding " $2\pi$ " to the argument does not change the number, we can also write $-i$ as modulus 1, argument $2\pi+ 3\pi/2= 7\pi/2$ . That gives, as a second root, $cos(7\pi/4)+ i sin(7\pi/4)= \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ .

In other words, the second root is just the negative of the first root, just as you learned way back in basic algebra!

**Ragnarok** · January 14th 2013, 05:31 PM

Thank you everyone for your input!

**zaidalyafey** · January 17th 2013, 10:28 AM

Originally Posted by Ragnarok

This is not really a calculus question but I am taking a course in basic complex analysis and this is from the first chapter.

Show that the indicated numbers satisfy the given equation and find an additional solution, $z_2$ .

$z^2+i=0, z_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

Obviously you can plug $z_1$ in and get a true statement, but how do you go about finding an additional solution?

$\text{if }z_1 \text{ and } z_2 \text{ are roots then we can write the polynomial as : }$

$(z-z_1)(z-z_2)=0 \Rightarrow \, \, z^2- z_1z-z_2z +z_1z_2 = 0 \, \Rightarrow \,\, z^2-(z_1+z_2)z +z_1z_2=0$

$\text{so comparing this to the original : }z_1+z_2 =0 \, \,\text{then }\fbox{ z_2=-z_1}$

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