Complex Analysis - Solutions of the Equation

**tarheelborn** · August 31st 2011, 06:32 PM

I need to find all the solutions of the equation
$z^2-4iz-4-2i=0$ .
I know that I can use the quadratic equation and then use a partial result of that to solve for the roots. However, when I use the quadratic equation, I come up with an answer that just doesn't make sense to me, i.e. $4i +- \sqrt{2i}$ . I am not sure where to go from here. I have gone over my calculations several times and this is just what I get... I will appreciate your help. (Sorry, I can't remember how to do the plus/minus sign). Thanks.

**mr fantastic** · August 31st 2011, 06:41 PM

Originally Posted by tarheelborn

I need to find all the solutions of the equation
$z^2-4iz-4-2i=0$ .
I know that I can use the quadratic equation and then use a partial result of that to solve for the roots. However, when I use the quadratic equation, I come up with an answer that just doesn't make sense to me, i.e. $4i +- \sqrt{2i}$ . I am not sure where to go from here. I have gone over my calculations several times and this is just what I get... I will appreciate your help. (Sorry, I can't remember how to do the plus/minus sign). Thanks.

I think it will be $2i \pm \sqrt{2i}$ . You can use DeMoivre's Theorem to get the two square roots of i in polar form and then convert to cartesian form.

**tarheelborn** · August 31st 2011, 06:58 PM

Originally Posted by tarheelborn

I need to find all the solutions of the equation
$z^2-4iz-4-2i=0$ .
I know that I can use the quadratic equation and then use a partial result of that to solve for the roots. However, when I use the quadratic equation, I come up with an answer that just doesn't make sense to me, i.e. $4i +- \sqrt{2i}$ . I am not sure where to go from here. I have gone over my calculations several times and this is just what I get... I will appreciate your help. (Sorry, I can't remember how to do the plus/minus sign). Thanks.

Thanks so much, but I really just need to know how to do the quadratic equation part. I think I can do the de Moivre's part if I can get that far. Silly, but that's the way it is. Sorry I posted in the wrong forum; I am really tired.

**mr fantastic** · August 31st 2011, 09:39 PM

Originally Posted by tarheelborn

Thanks so much, but I really just need to know how to do the quadratic equation part. I think I can do the de Moivre's part if I can get that far. Silly, but that's the way it is. Sorry I posted in the wrong forum; I am really tired.

I'd assumed that you could do the 'quadratic equation part' and had just made a careless error. Please post all your working and say where you get stuck.

**Prove It** · August 31st 2011, 09:56 PM

Originally Posted by tarheelborn

I need to find all the solutions of the equation
$z^2-4iz-4-2i=0$ .
I know that I can use the quadratic equation and then use a partial result of that to solve for the roots. However, when I use the quadratic equation, I come up with an answer that just doesn't make sense to me, i.e. $4i +- \sqrt{2i}$ . I am not sure where to go from here. I have gone over my calculations several times and this is just what I get... I will appreciate your help. (Sorry, I can't remember how to do the plus/minus sign). Thanks.

Either complete the square or use the Quadratic Formula.

**HallsofIvy** · September 1st 2011, 03:49 AM

Actually, you did NOT make an error, you simply have not finished the problem. You want to write the solutions in the "standard" form "a+ bi".

Use the quadratic formula (not "quadratic equation") $\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ with a= 1, b= 4i, and c= -4- 2i.

$\frac{-4i\pm\sqrt{(-4i)^2- (4)(1)(-4- 2i)}}{2(1)}= \frac{-4i\pm\sqrt{-16+ 16+ 8i}}{2}= \frac{-4i\pm\sqrt{8i}}{2}$
$= -2i+ \sqrt{2i}$ (which is what you got) where I have dropped the " $\pm$ " since I will be using the two roots of 2i.

"2i" is at (0, 2) in the complex plane, 90 degrees from the real axis, so its square roots are at 45 and -45 degrees, still distance 2 from the origin. That is, they are $\sqrt{2}(1+ i)$ and $\sqrt{2}(1- i)$ . The solutions to the equation are
$-2i+ \sqrt{2}(1+ i)= \sqrt{2}+ (\sqrt{2}- 2)i$ and $-2i+ \sqrt{2}(1- i)= \sqrt{2}- (\sqrt{2}+ 2)i$ .

**mr fantastic** · September 1st 2011, 01:52 PM

Originally Posted by HallsofIvy

Actually, you did NOT make an error, you simply have not finished the problem. You want to write the solutions in the "standard" form "a+ bi".

Use the quadratic formula (not "quadratic equation") $\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ with a= 1, b= 4i, and c= -4- 2i.

$\frac{-4i\pm\sqrt{(-4i)^2- (4)(1)(-4- 2i)}}{2(1)}= \frac{-4i\pm\sqrt{-16+ 16+ 8i}}{2}= \frac{-4i\pm\sqrt{8i}}{2}$
$= -2i+ \sqrt{2i}$ (which is what you got) where I have dropped the " $\pm$ " since I will be using the two roots of 2i.

"2i" is at (0, 2) in the complex plane, 90 degrees from the real axis, so its square roots are at 45 and -45 degrees, still distance 2 from the origin. That is, they are $\sqrt{2}(1+ i)$ and $\sqrt{2}(1- i)$ . The solutions to the equation are
$-2i+ \sqrt{2}(1+ i)= \sqrt{2}+ (\sqrt{2}- 2)i$ and $-2i+ \sqrt{2}(1- i)= \sqrt{2}- (\sqrt{2}+ 2)i$ .

The OP had 4i. (You have -2i !?). If the OP had read my post properly, s/he would have seen that it's 2i. As I said over a day ago, a careless error made by the OP.

However, the OP has marked the thread as solved and Thanked HoI and generally seems perfectly happy to swap one careless error for another, so who am I to rock the boat ....

**tarheelborn** · September 2nd 2011, 05:17 PM

Oops........ Thank you.

**tarheelborn** · September 3rd 2011, 04:25 PM

Originally Posted by mr fantastic

The OP had 4i. (You have -2i !?). If the OP had read my post properly, s/he would have seen that it's 2i. As I said over a day ago, a careless error made by the OP.

However, the OP has marked the thread as solved and Thanked HoI and generally seems perfectly happy to swap one careless error for another, so who am I to rock the boat ....

What is an OP?

**SammyS** · September 3rd 2011, 06:21 PM

Originally Posted by tarheelborn

What is an OP?

Original Poster or Original Post .

**tarheelborn** · September 3rd 2011, 06:49 PM

Thank you. I feared it was something negative.

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