I know this is wrong but I don't know why. In the set of complex numbers: $\sqrt1 = \sqrt{i^2\cdot i^2} = i\cdot i = -1$ What is wrong with this?
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Possible Duplicate:          I was thinking on the following line of thoughts: $1 =  \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$  Of course this is not true, but I was wondering which step in this 'line of thoughts' is forbidden to make?   Thanks for the explanation.
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If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?     Is it used or considered often in mathematics? How is it notated?
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$z$ is a complex number, by the way.  I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.  One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation \vert z + 1/z \vert^2 = a^2. These parts yield the constraints:  $2xy(1+1/(x^2+y^2)^2)=0$ ;   $x^2-y^2+\frac{2(x^2-y^2)}{x^2+y^2}+\frac{(x^2-y^2)}{(x^2+y^2)^2}=a^2 $  I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.  What can I do? Any hints on how to approach this?
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Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.  But when I checked Wolfram Alpha for , real it tells me it doesn't exist.  I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.
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$$\sum_{i=1}^n i^2 C^i_n = n(n+1) \cdot 2^{n-2}$$ I could use some help with proving that this equation stands. Should I try to use mathematical induction?
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Are they equal?      -5 = $\sqrt{(-5)^2}$
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I typed $\pi$ into Wolfram Alpha and in the short list of definitions there appeared  $$ \pi = -i \log(-1)$$  which really bothered me. Multiplying on both sides by $2i$:  $$ 2\pi i = 2 \log(-1) = \log(-1)^2 = \log 1= 0$$  which is clearly false.  I guess my error is $\log 1 = 0$ when $\log$ is complex-valued.  I need to use $1 = e^{2 \pi i}$ instead.  So my question: is it correct for WA to say $\pi = -i \log(-1)$?  Or should be they specifying "which" $-1$ they mean?  Clearly $ -1 = e^{i \pi}$ is the "correct" value of $-1$ here.
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Express $-1+i$ in exponential form.  My attempt so far  Let $z=-1+i$  $$r=|z|=\sqrt2$$  $$\theta=\tan^{-1}(-1)=-\frac{\pi}{4}$$  Now, this is where I go wrong (I don't know why it's wrong!):  So in exponential form: $-1+i=\sqrt2 e^{-i\pi/4}$  According to the solutions, $\theta=3\pi/4$.  Thanks in advance
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Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.
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Probably dumb question, but I would like to ask it anyway.  I was to solve this equation:  $z^4 = -16$  At first glance, the way to solve it would be (as any other equation):  $z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$  $z^2 = 4i \lor z^2 = -4i$  $z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$  However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number?
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I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.  But since $\sqrt x = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?  Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?  How do I reconcile all this? What rules am I not aware of?  Edit: Since someone voted to close my question, I should probably explain the difference between my question and , regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there's any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I'm asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)
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While teaching complex numbers for those who are encounter for the first time (usually 10th grader and 11th grader), I get the question like      "Can even squared number give negative results? How come there be square root of negative numbers? How much length does $i \; \mathrm{cm}$ represent in meter scale?"     in every year and in every section. I have been answering      "Well, this is just a construction such that it gives solution to equation $x^2 + 1 = 0$ which must have two solution by use of Fundamental theorem of algebra but since we do not have that in Real numbers, we are forced to construct $i$ that way." Then I go to history accepting the difficulty of some numbers like $\sqrt 2 $ by ancient  (just a red herring). Then I continue "Natural number is used to represent Money while Real number is used to measure length, complex number, while not representing length or weight, it my represent some other things than money or length that has applications in physics, math and engineering and so it is valid to take construct a symbol $i^2 = -1$."   I don't know if I am speaking too much out of my limited knowledge. What is the best way this knucklehead to explain those knucklehead who have been brainwashed in secondary and lower-secondary school that you cannot take square root of negative numbers, while still being correct and make them accept it easily?  EDIT: I have last question. Are some problems that is impossible to solve without use of Complex Numbers or all problems involving complex numbers have other workarounds?  EDIT-ADDED:: What would mathematics be like without the use of complex numbers? Perhaps, if I mention this clearly, maybe it would be helpful for students. One example is given by @Semiclassical on the comments in the first answer. What other things cannot be absolutely solved without the use of complex numbers?
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Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
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Consider the two imaginary numbers $i$ and $-i$. Is there any fundamental difference between the two of them? If I take the field $\mathbb{C}$ and apply the map $a + bi \mapsto a - bi$ does the image end up meaningfully different from the field I started with? Or when we write out complex numbers are we arbitrarily choosing which of the non-real solutions to $z^4 = 1$ to call $i$ and which to call $-i$?
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Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$?  Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.
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So, my friend show me prove that $1=-1$ by using this way:  $$1=\sqrt{1}=\sqrt{(-1)\times(-1)}=\sqrt{-1}\times\sqrt{-1}=i\times i=i^2=-1$$  At first sight, I stated "No, $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ is valid only for $a,b\in\mathbb{R}$ and $a,b\geq0$"  But, I remember that $\sqrt{-4}=\sqrt{4}\times\sqrt{-1}=2i$ which is true (I guess).  Was my statement true? But, $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ is also valid if one of a or b is negative real number. Why is it not valid for a dan b both negative? If my statement was wrong, what is wrong with that prove?
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I think I've reached a solution for the above question, I'd like to know if my answer makes sense, and my process follows the appropriate steps, I'm rather new to working in the complex space.   Let $z = r(\cos\theta + i\sin\theta)$. Then:  $z^{n-1} = \overline z$   $z^n = z\overline z$, where $\overline z = r(\cos\theta - i\sin\theta)$. Then for $z\overline z = r^2(\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) = r^2$, thus we get:  $z^n = r^2$  $r^n(\cos n \theta + i\sin n\theta)  = r^2 \implies r^{n-2}(\cos n \theta + i\sin n\theta) = 1$. This gives us the set:  $\{z = r(\cos\theta + i\sin\theta) : r^{n-2}(\cos n \theta + i\sin n\theta) = 1\}$  Continuing as mentioned in the comments we can solve for $r$ and $\theta$. Rewriting we get (from here let $cis\theta = \cos\theta + i\sin\theta$):  $r^{n-2}cisn\theta = 1cis0.$ Clearly, $r^{n-2} = 1$ and since $r \geq 0$ then $r = 1$ for $n \geq 3$. And then:  $n\theta = 0 + 2\pi k, k = 0, 1, 2, 3, \dots n-1$  So $\theta  = \frac{2\pi k}{n}, k = 0, 1, 2, \dots n -1 $. This gives us the set:  $\{z = cis\theta : \theta = \frac{2\pi k}{n}, k = 0, 1, 2, \dots n -1, n \geq 3\}$