September 7th, 2017, 05:35 AM  #31  
Global Moderator Joined: Dec 2006 Posts: 18,166 Thanks: 1424  Quote:
As this is a basic algebra problem that happens to involve complex numbers, it doesn't belong in the complex analysis section.  
September 7th, 2017, 05:50 AM  #32 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,042 Thanks: 2344 Math Focus: Mainly analysis and algebra  He's not wrong. But for some reason you refuse to accept that your method that took several iterations and a heap of working isn't better. This despite the fact that your approach apparently still hasn't produced the right answer  although I haven't been checking on that. A similar this is/was going on in the recent limit thread. This is a common thread in your posts: an inability to accept when you are incorrect or to accept that ideas contrasting to your own might be better, or at least no worse. 
September 7th, 2017, 09:32 PM  #33 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
Review of skipjack's solution, post #4 1) $\displaystyle z^{2}+2(1+i)z+2=0$ The form suggests a step I wouldn't have taken assuming it just leads to quadratic formula: 2) $\displaystyle [z+(1+i)]^{2}=z^{2}+2(1+i)z+(1+i)^{2}$ Substituting $\displaystyle z^{2}+2(1+i)z=(z+1+i)^{2}(1+i)^{2}$ into 1) gives: $\displaystyle (z+1+i)^{2}=2(i1)$ Now comes the tricky part. Let 3) $\displaystyle a+bi=z+1+i$. and then 4) $\displaystyle (a+bi)^2=2(i1)$ and square a+bi rather than trying to find $\displaystyle [2(1i)]^{\frac{1}{2}}$ Solve for a and b by equating real and imaginary parts of 4). Then find z from 3). At this point the actual algebra is elementary and, in my opinion, a waste of time once you understand how you got here, but it gives for z: $\displaystyle z=1+\sqrt{\sqrt{2}1}+i\left ( 1+\sqrt{\sqrt{2}+1} \right )$ $\displaystyle z=1\sqrt{\sqrt{2}1}+i\left ( 1\sqrt{\sqrt{2}+1} \right )$ Comment: It's nice because it let's you solve for z without having to find $\displaystyle [2(1i)]^{\frac{1}{2}}$. The actual calculation is much simpler algebra than letting z=a+bi and substituting into OP. Unfortunately, I missed the "tricky" part and didn't understand this proof, and assumed it couldn't be right because you couldn't do it without taking square root of a complex number somewhere (assuming it didn't factor into a perfect square), but the same procedure works for the general complex quadratic. So it isn't just a matter of elementary algebra, aside from the trick As for the "simple" comments, no comment. 
September 8th, 2017, 11:12 AM  #34 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
If you get to a point where $\displaystyle (z+z_{1})^{^{2}}=z_{2}$, is there any advantage in letting $\displaystyle z+z_{1}=a+bi$ and writing $\displaystyle a^{2}b^{2}+2bi=z_{2}$ and solving for a and b, over simply writing $\displaystyle z=z_{1}+z_{2}^{\frac{1}{2}}$ which is already the answer in an accepted form? $\displaystyle z_{2}=re^{i \theta }, z_{2}^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i \frac {(\theta+2k\pi)}{2}}, k=0,1 $ What if we had $\displaystyle (z+z_{1})^{3}=z_{2}$? Should you think, then $\displaystyle (a+bi)^{3}=a^{3}+a^{2}biab^{2}b^{3}i+2abi2b^{2}=Rez_{2}+Imz_{2}i$ and solve for a and b, as opposed to $\displaystyle z_{2}=re^{i\theta}, z_{2}^{\frac{1}{3}}$ $\displaystyle =r^{\frac{1}{3}}e^{i\frac{(\theta+2k\pi)}{3}}, k=0,1,2$ The point being, the "trick" is not a preferred solution to $\displaystyle z^{\frac{1}{2}}$ as implied in the thread, but it is an easier calculation than sustituting a+bi into OP, which turns out to be a very awkward calculation. In other words, the "trick" is not a solution at all to OP, it is just a clever algebraic shortcut seen in the course of the solution. 
September 8th, 2017, 11:18 AM  #35 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,042 Thanks: 2344 Math Focus: Mainly analysis and algebra 
Perhaps you should read the original question properly. Why are you so desperate to show that your idea is the only correct one? There are always multiple ways to reach a result. Yes, you can use $\sin\frac38\pi$ and $\cos\frac38\pi$ (or whatever) in your answer, but it's not very informative. Last edited by v8archie; September 8th, 2017 at 11:23 AM. 
September 8th, 2017, 11:36 AM  #36 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  
September 8th, 2017, 12:05 PM  #37 
Global Moderator Joined: Dec 2006 Posts: 18,166 Thanks: 1424 
For the posted problem, $\theta$ can be found quite easily, zylo, but you omitted to explain how in your post. You would later need to calculate $\cos(\theta/2)$ and $\sin(\theta/2)$, and you omitted to explain how to do that. These steps are required, as the problem specifies that the answer is to be given in the form a + bi. You could add these explanations, of course, but they considerably lengthen your method, so that it's no longer "simple".

September 11th, 2017, 06:22 AM  #38  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  Quote:
1) $\displaystyle z^{2}+k_{1}z+k_{2}=0$ and $\displaystyle z,k_{1},k_{2}$ complex. "Completing the square" $\displaystyle (z+k_{1}/2)^{2}=z^{2}+k_{1}z+k_{1}^{2}/4$ Substitute $\displaystyle z^{2}+k_{1}z$ into 1). Then $\displaystyle (z+k_{1}/2)^{2}=k_{2}+k_{1}^{2}/4=c+di$ Let $\displaystyle a+bi=z+k_{1}/2$ $\displaystyle (a+bi)^{2}=c+di$ $\displaystyle a^{2}b^{2}=c$ $\displaystyle 2ab=d,d=b/(2a))$, Then $\displaystyle 4a^{4}4a^{2}cd^{2}=0$ $\displaystyle t^{2}ctd^{2}/4=0$ $\displaystyle t=[c\pm\sqrt{c^{2}+d^{2}}]/2$ which is always + or . Then $\displaystyle a=\pm\sqrt{t}, b=d/2a, z+k_{1}/2=a+bi$ give both roots. It's a viable procedure. Can't find it on Google or a text book. Is it in new texts, is it "out there," or did you just "see" it? skipjack, why not post the procedure in a new thread for general availability?  
September 11th, 2017, 08:06 AM  #39 
Global Moderator Joined: Dec 2006 Posts: 18,166 Thanks: 1424 
You slipped up towards the end. The procedure can be found by use of Google. It's described in the textbook Maths: A Student's Survival Guide (subtitled A SelfHelp Workbook for Science and Engineering Students) by Jenny Olive (ISBN: 9780521017077) and it's also described here. 
September 11th, 2017, 10:17 AM  #40 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90 
So why is the general solution for a cubic equation with real coefficients valid when the coefficients are complex? Without duplicating real solution step by step? https://en.wikipedia.org/wiki/Cubic_function (Same principle works for OP) EDIT Or a quartic. Last edited by zylo; September 11th, 2017 at 10:20 AM. 

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