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August 17th, 2017, 03:12 PM   #1
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Math Focus: differentiation
Question solve the eqn z^2+2(1+i)z+2=0,giving each result in the form a+bi.

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August 17th, 2017, 03:47 PM   #2
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What have you tried
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August 31st, 2017, 05:37 AM   #3
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What principle can you apply?

Complex numbers are a field (satisfy same axioms as real numbers) so you can use the quadratic formula. Then you have to do some algebra to put the two answers in a+bi form. In particular, you need to know $\displaystyle z^{\frac{1}{2}}$
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August 31st, 2017, 02:40 PM   #4
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$z^2 + 2(1 + i)z + 2 = 0$ implies $(z + 1 + i)^2 = (1 + i)^2 - 2 = 2i - 2$.
Let $(a + bi)^2 = 2i - 2$, where $a$ and $b$ are real.
That implies $a^2 - b^2 = -2$ and $ab = 1$.
Eliminating $b$, $a^2 - 1/a^2 = -2$ and so $(a^2 + 1)^2 = 2$.
Hence $a = \pm\sqrt{-1 + √2}$.
Similarly, $b = \pm\sqrt{1 + √2}$, where $a$ and $b$ have the same sign.
Hence $z = -1 + \sqrt{-1 + √2} + (\sqrt{1 + √2} - 1)i$ or $-1 - \sqrt{-1 + √2} - (\sqrt{1 + √2} + 1)i$.
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September 1st, 2017, 09:01 AM   #5
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Let z=a+bi, substitute into OP, and set real and imaginary parts to 0 to solve for a and b.

Straight-forward, no gimmicks.
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September 1st, 2017, 10:22 AM   #6
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No mention of the irrelevant $z^{\frac12}\!$ now. Have you tried to solve the equations?
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September 1st, 2017, 11:08 AM   #7
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Quote:
Originally Posted by skipjack View Post
No mention of the irrelevant $z^{\frac12}\!$ now. Have you tried to solve the equations?
I was just pointing out z was a field so you could use quadratic formula. I agree the algebra is tedious.

z=a+bi and substitute is the most straight-forward, but yes, the algebra is also tedious. I quit after a while- I don't like wading through algebra, and I always make mistakes because if it's tedious I'm thinking of other things while I do it.

Your solution is very nice except for the fact that you had to think of a simplifying substitution which would never occur to me, and which frankly I don't even understand.
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September 1st, 2017, 05:30 PM   #8
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Quote:
Originally Posted by skipjack View Post
No mention of the irrelevant $z^{\frac12}\!$ now. Have you tried to solve the equations?
$\displaystyle z^{2}+2(1+i)z+2=0$
$\displaystyle z=[-2(1+i) \pm \sqrt{4(1+i)^{2}-8}]/2$
$\displaystyle \sqrt{4(1+i)^{2}-8}=\sqrt{8}\sqrt{i-1}$
$\displaystyle i-1=\sqrt{2}e^{\frac{3\pi i}{4}+2k\pi i}$
$\displaystyle \sqrt{i-1}=2^{\frac{1}{4}} e^{\frac{3\pi i}{8}}, -2^{\frac{1}{4}}i$
$\displaystyle \sqrt{i-1}= 2^{\frac{1}{4}}(\cos60+i\sin60), -2^{\frac{1}{4}}i$
$\displaystyle \sqrt{8}\sqrt{i-1}=2^{\frac{3}{4}}(1+\sqrt{3}i),-2^{\frac{7}{4}}i$

And that's the problem:
$\displaystyle z=(-b\pm \sqrt{z_{1}})/2\\$

But $\displaystyle \sqrt{z_{1}}$ has two values, so z has four values, i.e., a complex quadratic has four roots.

Does anyone have a math program to solve this?

Last edited by skipjack; September 2nd, 2017 at 02:51 AM.
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September 2nd, 2017, 02:34 AM   #9
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Quote:
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But $\displaystyle \sqrt{z_{1}}$ has two values
$\sqrt{.}$ is a function $\mathbb{C} \to \mathbb{C}$, so $\sqrt{z}$ is just a single complex number for any $z \in \mathbb{C}$ (the principle square root of $z$).

However, if $v^2 = z_1$ then $v$ is one of the two square roots of $z_1$: $v = \sqrt{z_1}$ or $v = -\sqrt{z_1}$. Fortunately, by writing $\pm \sqrt{z_1}$ in your preceding working, you've already accounted for both square roots - there's no need to count them both again by saying $\sqrt{z_1}$ has two values!

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Originally Posted by zylo View Post
So z has four values, ie, a complex quadratic has four roots.
This is why it's actually a big problem that you've counted the same thing twice! Hopefully, if you take a step back, you'll see why this couldn't be true. How could a polynomial of degree $n$ over a field have more than $n$ roots in that field?
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Last edited by cjem; September 2nd, 2017 at 02:37 AM.
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September 2nd, 2017, 02:54 AM   #10
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Quote:
Originally Posted by zylo View Post
Let z=a+bi, substitute into OP, and set real and imaginary parts to 0 to solve for a and b.
Have you attempted to solve the equations that result from using that method?
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