Author Topic: wildareal's questions thread (Read 21215 times) Tweet Share

wildareal · « **on:** December 30, 2010, 03:58:48 pm »

If z=1+ i is one solution of an equation of the form z^4=a, where a E C, what are the other solutions?

Thanks.

kakar0t · « **Reply #1 on:** December 30, 2010, 04:17:50 pm »

Convert z=1+i to polar form and then add/subtract pi from the argument (90 degrees) to find the other solutions.

so

z=1+i -> sqrroot(2)cis(pi/4)
adding/subtracting pi gives us the other solutions

sqrroot(2)cis(3pi/4)
sqrroot(2)cis(-pi/4)
sqrroot(2)cis(-3pi/4)

kamil9876 · « **Reply #2 on:** December 30, 2010, 04:18:09 pm »

The other solutions are those that you get from rotating it by multiples of $\pi/2$ . So rotating it once gives $-1+i$ , rotating again gives $-1-i$ , rotating again gives $1-i$ , and rotating again...

(Draw this)

wildareal · « **Reply #3 on:** December 30, 2010, 10:49:19 pm »

The roots of the equation 2x^2 +6x +7=0 are f and g. The value of mod(f-g) is?

Thanks.

kamil9876 · « **Reply #4 on:** December 30, 2010, 11:11:52 pm »

quadratic formula gives:

$f= (-6 + \sqrt{-20})/4$

$g=(-6 - \sqrt{-20})/4$

$so |f-g|=|\sqrt{-20}/2|=|i\sqrt{20}/2|=\sqrt{20}/2=\sqrt{5}$

wildareal · « **Reply #5 on:** January 02, 2011, 11:12:56 am »

Express 1/(1+iroot(3)) in the form rcis(theta), where r>0 and -TT<theta≤TT.

Thanks.

jasoN- · « **Reply #6 on:** January 02, 2011, 11:52:25 am »

$\frac{1}{1+i\sqrt{3}} * \frac{1-i\sqrt{3}}{1-i\sqrt{3}}$

$\frac{1-i\sqrt{3}}{1-i^2*3}$

$\frac{1-i\sqrt{3}}{-2}$

$\frac{-1}{2} +\frac{i\sqrt{3}}{2}$

$r = |\frac{-1}{2} + \frac{i\sqrt{3}}{2}| = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$

$Arg(\frac{-1}{2} + \frac{i\sqrt{3}}{2}) = \pi - tan^{-1} (\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}) = \pi - tan^{-1} (\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

$\therefore \frac{1}{1+i\sqrt{3}} = 1cis(\frac{5\pi}{6})$

hope it's right

pi · « **Reply #7 on:** January 02, 2011, 01:18:34 pm »

Quote from: jasoN- on January 02, 2011, 11:52:25 am

$\frac{1-i\sqrt{3}}{1-i^2*3}$

$\frac{1-i\sqrt{3}}{-2}$

Shouldn't the denominator be 4, not -2... (I haven't done much complex stuff yet so I could be badly wrong...)

Russ · « **Reply #8 on:** January 02, 2011, 01:23:36 pm »

Quote

Shouldn't the denominator be 4, not -2... (I haven't done much complex stuff yet so I could be badly wrong...)

All the letters in this fancy maths confuse me, but as far as I can see, yes.

jasoN- · « **Reply #9 on:** January 02, 2011, 01:36:17 pm »

ah yes you are right i was thinking i^2 = 1
nevertheless the method is there

wildareal · « **Reply #10 on:** January 02, 2011, 01:36:40 pm »

Hey you have to rationalise it (thanks for the tip jasoN), then you get (1-iroot(3))/4

mod(z)=1/2

tan(theta)=root(3)/4 x 4
tan(theta)=root(3)

4th Quadrant:

Therefore -TT/3

z=1/2cis(-TT/3)

kamil9876 · « **Reply #11 on:** January 02, 2011, 03:27:17 pm »

much easier this way:

$1+i\sqrt{3}=2cis\pi /3$

Hence $\frac{1}{1+i\sqrt{3}}=\frac{cis0}{2cis(\pi/3)}=\frac{1}{2} cis(-\pi/3)$

In general, multiplying or dividing in polar form is much easier and that it is why it was easier to convert first.

pi · « **Reply #12 on:** January 02, 2011, 05:18:50 pm »

^^That's brilliant! Thanks for the tip!

wildareal · « **Reply #13 on:** January 02, 2011, 08:36:20 pm »

Find the real numbers k such that z=ri is the root of the equation:

z^4-2z^3 +11z^2-18z+18=0

Thanks.

kamil9876 · « **Reply #14 on:** January 02, 2011, 08:50:42 pm »

do you mean real numbers $r$ ?

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