Melody

If    Z^4=-5-3i                      Find z

First, there will be 4 forth roots because that is how it works.

I need to write -5-3i  in polar form.

The easiest way to tackly this in my opinion is to graph the point.   and then work out the distance r from 0+0i and the angle from the positive x axis.  You can memorize formulas if you want but I can never remember those.

r = sqrt(25+9) = sqrt(34)       ( I just used pythagoras's theroem from the pic)

This is in the 3rd quad so theta =[ 180+  atan(3/5)]       degrees

It is not allowing me to edit my LaTex but I have used r=6 through out this solution and it should have been

r= sqrt34.

Hence all  the final answers will be slightly out. 

so

\(-5-3i\;\;\;=6\;[cos(180+atan(0.6))\;\;+\;\;i\;sin(180+atan(0.6))]\\ \sqrt[4]{-5-3i}=\sqrt[4]{6}\;\angle\;\alpha\qquad where \;\;\alpha =\left[\frac{180+atan(0.6)+360 k}{4} \right]\qquad \mbox{For k=0,1,2 and 3}\\ \alpha =\frac{180+atan(0.6)+360 k}{4}\\ \alpha =\frac{180+atan(0.6)}{4} + \frac{360 k}{4} \\ \alpha =\frac{180+atan(0.6)}{4} + 90k \\ \alpha =\frac{180+30.96}{4} + 90k \\ \alpha \approx 52.74 + 90k \\ \alpha \approx52.74 \quad or \quad \alpha \approx52.74+90 \quad or \quad \alpha \approx 52.74+180 \quad or \quad \alpha \approx 52.74+270 \\ \alpha \approx52.74\quad or \quad \alpha \approx 180-37.26 \quad or \quad \alpha \approx 52.74+180 \quad or \quad \alpha \approx 360-37.26 \\ \)

\(-5-3i\;\;\;=6\;[cos(180+atan(0.6))\;\;+\;\;i\;sin(180+atan(0.6))]\\ \sqrt[4]{-5-3i}=\sqrt[4]{6}\;\angle\;\alpha\qquad where \;\;\alpha =\left[\frac{180+atan(0.6)+360 k}{4} \right]\qquad \mbox{For k=0,1,2 and 3}\\ \alpha \approx52.74\quad or \quad \alpha \approx 180-37.26 \quad or \quad \alpha \approx 52.74+180 \quad or \quad \alpha \approx 360-37.26 \\ cos(52.74)\approx 0.6054\qquad sin(52.43)\approx 0.7959\\ root1:\qquad \approx \sqrt[4]{6}[cos(52.74)+isin(52.74)] \\ root1:\qquad \approx 1.5651[cos(52.74)+isin(52.74)] \\ root1:\qquad \approx 1.5651[0.6054+0.7959i] \\ root1:\qquad \approx 0.95+1.25i \\ .....\\ root2:\qquad \approx 1.5651[cos(180-37.26)+isin(180-37.26)] \\ root2:\qquad \approx 1.5651[-cos(37.26)+isin(37.26)] \\ root2:\qquad \approx 1.5651[-0.7959+0.6054i] \\ root2:\qquad \approx -1.25+0.95i \\ .....\\ root3:\qquad \approx 1.5651[cos(52.74+180)+isin(52.74+180)] \\ root3:\qquad \approx 1.5651[-cos(52.74)-isin(52.74)] \\ root3:\qquad \approx 1.5651[-0.6054-0.7959i] \\ root3:\qquad \approx -0.95-1.25i \\ .....\\ root4:\qquad \approx 1.5651[cos(360-37.26)+isin(360-37.26)] \\ root4:\qquad \approx 1.5651[cos(37.26)-isin(37.26)] \\ root4:\qquad \approx 1.5651[0.7959-0.6054i] \\ root4:\qquad \approx 1.25-0.95i \\ \)

\(root1:\qquad \approx 0.95+1.25i \\ .....\\ root2:\qquad \approx -1.25+0.95i \\ .....\\ root3:\qquad \approx -0.95-1.25i \\ .....\\ root4:\qquad \approx 1.25-0.95i \\ \)

**    It will not let me edit my LaTex but each of these is a possible value of z

Here is a diagram showing -5-3i and also showing its 4 forth roots.    

Hi Complex,

It is nice to meet you :)

The table shows estimates of computer storage. A CD - ROM hold 700 MB. A DVD - ROM hold 4.7 GB. Estimate how many times more storage a DVD has than a CD. Explain how you found you answer. 1 kilobyte(KB) = 1000 bytes 1 megabyte(MB) = 1 million bytes 1 gigabyte(GB) = 1 billion bytes

CD     700MB

DVD   4.7GB = 4.7*1000 MB  = 4700 MB

4700/700 = 47/7   = approx  6.71

So  a DVD hold approx  6.71 times the amount of data that a CD holds

ln(x^2-x-20/(x+6)^4)^1/3

\(ln(x^2-x-20/(x+6)^4)^{1/3}\\ =ln\left(\frac{x^2-x-20}{(x+6)^4}\right)^{1/3}\\ =\frac{1}{3}\left[\ln\left(\frac{x^2-x-20}{(x+6)^4}\right)\right]\\ =\frac{1}{3}\left[ln(x^2-x-20)-ln(x+6)^4\right]\\ =\frac{1}{3}\left[ln(x^2-x-20)-4ln(x+6)\right]\\ =\frac{\ln(x^2-x-20)-4ln(x+6)}{3}\\ \)

-5*+3* if *=-6

\(-5^{-6}+3^{-6} =\frac{-1}{15625}+\frac{1}{729}\approx 1.307742112 \times 10^{-3}\)

8x - 2(3x - 4 + 2y) - 12x

=8x + 2(-3x + 4 - 2y) - 12x

=8x  -6x + 8 - 4y - 12x

=8x  -6x - 12x + 8 - 4y 

= -10x + 8 - 4y 

\(-1-1/2x(-7/14-2)\\ =-1-\frac{x}{2}(-\frac{7}{14}-2)\\ =-1-\frac{x}{2}(-\frac{1}{2}-\frac{4}{2})\\ =-1\quad -\frac{x}{2}(-\frac{5}{2} )\\ =-1\quad +\frac{5x}{4}\\ =-\frac{4}{4}\quad +\frac{5x}{4}\\ =\frac{5x-4}{4}\)

Yes that is right.

Supplementary is add to 180 and complementary is add to 90.

It is easy to remember which is which because 2 times 90 = 180 and 2 c's on top of one another, make an S.

Does that make sense?

anyway...

(4x+9)+(6x+3) = 180

4x+9+6x+3 =180

10x+12 =180

10x=168

x=16.8

I can see straight off that 6x +3 is going to be bigger  than 4x+9

so

the big one will be  6*16.8 +3 = 103.8 degrees   ( 103 degrees and 48 minutes)

check

the little one will be  4*16.8+9 = 76.2

103.8+76.2=180   great :))

Well, can you tell me what supplementary means?

Yes CPhill,  You have made many very good observations here.

You know, fingers in pies is something I expect Sir Phill La-Tex is quite familiar with.

Chimps so like to put their fingers in their pies ..........

I wonder how that formula might go.

\(V=\displaystyle \int_0 ^{chimp\; digit\;l ength}\left(\pi r^2h-6\frac{\delta\pi}{\delta r}\right)dh\)

I had to subtract the 6 partial pies because Sir La-Tex had 5 of his friends over.

There was Lancelotlink, .....   I can't remember who the others were.  

There was that brilliant female one, I have forgotten her name.....    

Anyway, all chimps look the same to me.     

Thanks Bertie,

I was wondering why you had become so scarce.

I wonder if that is why so many of our regulars have stopped appearing? 

I just thought that they did not like the new forum.

I will let Andre know that this is an ongoing problem and maybe he will fix it.  :)

Yes I did have a good trip to England, thanks :)