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A Complex Problem

February 13, 2012 in Mathematics Posts | Tags: , , , , , , , | Leave a comment

I wrote this note in 2008 to introduce complex numbers. It is posted for the Math Teachers at Play blog carnival.

What is the {\sqrt[4]{-16}}?

In the set of Real Numbers, {\mathbb{R}}, the solution is undefined. We must consider the set of Complex Numbers, {\mathbb{C}}. Complex numbers are numbers of the form of {a+bi}, where {a} and {b} are real numbers: {a,b\in\mathbb{R}} and {i} is the imaginary unit. Note {i} is defined by {i^{2}=-1} or equivalently {i=\sqrt{-1}}.

Thus, we have:

\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & (-16)^{1/4}\\ & = & (-1\cdot16)^{1/4}\\ & = & (-1\cdot2^{4)^{1/4}}\\ & = & (-1)^{1/4}\cdot2^{4/4}\\ & = & (-1)^{1/4}\cdot2\\ & = & ((-1)^{1/2})^{1/2}\cdot2\\ & = & (\sqrt{-1})^{1/2}\cdot2\\ & = & i^{1/2}\cdot2\\ & = & 2\sqrt{i}.\end{array}

We now have {\sqrt[4]{-16}=2\sqrt{i}}. So we must ask "What is the {\sqrt{i}}?". To evaluate {\sqrt{i}} we will use Euler’s formula, {e^{ix}=\cos\left(x\right)+i\sin\left(x\right).} So by substituting {x=\frac{\pi}{2}}, giving

\displaystyle \begin{array}{rcl} e^{i(\frac{\pi}{2})} & = & \cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\\ & = & 0+i(1)\\ & = & i.\end{array}

Note we have isolated {i} and now arrived at the following equality: {i=e^{i(\frac{\pi}{2})}}. Taking the square root of both sides gives

\displaystyle \begin{array}{rcl} \sqrt{i} & = & \sqrt{e^{i(\frac{\pi}{2})}}\\ & = & e^{i(\frac{\pi}{2})(\frac{1}{2})}\\ & = & e^{i(\frac{\pi}{4})}.\end{array}

Thus, we have now have {\sqrt{i}=e^{i(\frac{\pi}{4})}} and following through application of Euler’s formula to {x=\frac{\pi}{4}}, gives

\displaystyle \begin{array}{rcl} \sqrt{i} & = & e^{i(\frac{\pi}{4})}\\ & = & \cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\\ & = & \frac{1}{\sqrt{2}}+i(\frac{1}{\sqrt{2}})\\ & = & \frac{\sqrt{2}}{2}+i(\frac{\sqrt{2}}{2})\\ & = & \frac{\sqrt{2}+\sqrt{2}i}{2}.\end{array}

So going back to the original problem and substituting {\sqrt{i}=\frac{\sqrt{2}+\sqrt{2}i}{2}}, gives

\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & 2\sqrt{i}\\ & = & 2(\frac{\sqrt{2}+\sqrt{2}i}{2})\\ & = & \sqrt{2}+\sqrt{2}i.\end{array}

Hence, the solution, {\sqrt[4]{-16}=\sqrt{2}+\sqrt{2}i} , is a complex number!

Zeros of a Polynomial

November 24, 2011 in Mathematics Posts | Tags: , , , , , , , , , | 1 comment

Proposition 1 Prove {\sqrt{2}} is irrational.

Here is a proof using a traditional method (See Euclid’s Elements Book X which incorporates Theatetus work on incommensurable numbers. It includes a proof that {\sqrt{2}} is irrational (Proposition 22), and ends with a proof that there are infinitely many distinct irrational numbers (Proposition 115): Read the rest of this entry »

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