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Square root of a complex number

SQUARE ROOT OF A COMPLEX NUMBER

The solution of the equation $$z^2=a+ib$$ is given in two forms.

In terms of $a,b$,
$$ z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} + i {b\over |b|}\sqrt{{ \sqrt{a^2+b^2}-a \over  2}}\right)$$

In terms of $w=a+ib$,
$$ z= \pm \left(\sqrt{{ |w|+\Re(w) \over  2}} + i{\Im(w)\over |\Im(w)|} \sqrt{{ |w|-\Re(w) \over  2}}\right)$$

A crazy person could just check this by squaring it! you should get $w=a+ib$ back.

=========================== WORKING BELOW

Suppose we wish to find the square root of a complex number $w=a+ib$.

Some notation we use later regarding real and imaginary parts of the complex number.

$\Re(w)=a$
$\Im(w)=b$
${x\over |x|}$   gives the sign of $x$, $+1$ or $-1$.

Let $z=x+iy$, and of course keep in mind that we use $i^2=-1$.


$$\begin{array}{ll} z^2 &=a+ib\\ (x+iy)^2 &= a+ib\\x^2-y^2+i2xy &= a+ib  \end{array}$$

Therefore equating real and imaginary parts we have two equations:

$x^2-y^2=a$ and $2xy=b$. 

$x^2-\displaystyle{b^2\over 4x^2}=a$

$4 x^4-b^2=4 x^2 a$

$4x^4-4a x^2 -b^2 = 0$

$x^2 = {4a \pm \sqrt{16a^2-4\times 4(-b^2)} \over 8}$

$x^2 = {4a \pm \sqrt{16a^2+16b^2} \over 8}$
$x^2 = {a \pm \sqrt{a^2+b^2} \over 2}$

$x^2 = {a + \sqrt{a^2+b^2} \over 2}$
 since $x^2\ge 0$ needs to hold.

$\therefore x = \pm \sqrt{{a + \sqrt{a^2+b^2} \over 2}}$

$\therefore y= \pm {b\over 2} \sqrt{{2 \over a+\sqrt{a^2+b^2} }}$

rationalise the denominator and bring the '$b/2$' inside the square root symbol by keeping in mind that $b=sign(b)\times |b|$.

$\therefore y= \pm sign(b) \sqrt{{b^2\over 4}\,{2 \over a+\sqrt{a^2+b^2} }  \times { a-\sqrt{a^2+b^2} \over a-\sqrt{a^2+b^2}}}$

Finally in terms of $a$ and $b$ we get

$$ z= \pm \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} \pm i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over  2}}$$

or in terms of $w$

$$ z= \pm \sqrt{{ |w|+\Re(w) \over  2}} \pm i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over  2}}$$

The $\pm$ means you either take both as $+$ or both as $-$. :-)

Or with a big bracket but looks more cumbersome ..

So in terms of $a,b$,
$$ z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over  2}} + i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over  2}}\right)$$

Or in terms of $w$,
$$ z= \pm \left(\sqrt{{ |w|+\Re(w) \over  2}} + i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over  2}}\right)$$

A crazy person could just check this by squaring it! you should get $w=a+ib$ back.

====================
Postscript Note:
Apparently it can also be shown that

$$\sqrt{z}=\sqrt{|z|}\,{ z+|z|\over \left| z + |z| \right| }$$

where $\sqrt{z}$ means the principal square root, which is where the real part is positive.

You can prove this formula by taking $z=x+iy$ and brute force it for example :-), enjoy!!

New largest prime discovered $2^{74\,207\,281}-1$

New largest prime discovered 

$$\Huge2^{74\,207\,281}-1$$

Recall a prime number is one which is only divisible by itself and $1$ and the smallest prime numbers are $2,3,5,7,11,13,17,...$.

The new largest prime number has got over $22$ million digits! that's right!
The number $1\,000\,000$ has $7$ digits! Can you imagine a number with $22$ million digits!
A book may have say $2000$ characters to a page, so that's about $22\,000\,000\div 2000=11\,000$ pages just to write out the number long hand!

PHYS.ORG

Euler's Formula for Simple Polyhedras $F+V-E=2$

Definition: 
A SIMPLE POLYHEDRON is a polyhedron which has no holes.

THEOREM:
Euler's Formula for Simple Polyhedra. 
If $V$ is the number of vertices, $E$ the number of edges, and $F$ the number of faces of a simple polyhedron, then $$F+V-E=2$$

Note: Most articles will quote this result for convex polyhedrons only, but it holds for those concave polyhedra without holes also.

NUMBERS AND THE UNIVERSE.

NUMBERS AND THE UNIVERSE.

Here is a great reference for the number of stars in our galaxy ($\ge 100$ billion stars) and the number of galaxies in the observable Universe ($\ge 100$ billion galaxies) and my favourite,, dark energy and dark matter makes up $95\%$ of the Universe. The rest is only $ 5\%$, that is the observable universe, all the stars, planets,... constitute only $5\%$ of the Universe! Well, that's according to the latest in science at NASA..

http://nasa.tumblr.com/post/136762377389/7-facts-that-will-make-you-feel-very-small

Bit of Fun. Patterns in multiplications.

Bit of Fun. Patterns in multiplications.
I saw this online somewhere. Instead oif $8$ you could put another single digit number but most give boring results. Except $9$ which gives an impressive regular pattern, see below.

$\Large 1\times 9+1=10$

$\Large 12\times 9+2=110$

$\Large 123\times 9+3=1110$

$\Large 1234\times 9+4=11110$

$\Large 12345\times 9+5=111110$

$\Large 123456\times 9+6=1111110$

$\Large 1234567\times 9+7=11111110$

$\Large 12345678\times 9+8=111111110$

$\Large123456789\times 9+9=1111111110$

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