Fredrick's questions

[fredrick]

ok, here we go.

Q1) One of the complex solutions to z^5=-a, where a is a positive real constant, is .
One of the other solutions is a real number and is equal to:
A)
B)
C)
D)
E)

More to come!

[Mao]

B

all root of follow a simple pattern: they differ by radians

that is, all roots here differ by , so your roots are

all non-zero real number have the arguments (positive real) or (negative real) [remember the definition of cis, cos + i*sin. Where the sin term evaluates to 0, the number is a real number]

in this case, only satisfies this. Hence, the correct option is B

[shinny]

Fastest way to work that out is that since its a fifth power polynomial, there will be 5 solutions in total definitely. Also, since these solutions will be spaced out equally, they will be separated by an angle of 2pi/5 from one another. So from a^(1/5)cis(pi/5), add and subtract 2pi/5 until the solutions start repeating. Since you're looking for a real number, the argument would have to be pi or 0. 0 is not possible to obtain from adding multiples of 2pi/5, but pi is (add 2 sets of 2pi/5). So your answer is a^(1/5)cis(pi) i.e. -a^(1/5)

p.s. sorry for lack of latex - I'm too slow at it and I find it hard to do worded explanations within it <_<

edit: oh mao's got his explanation up too now. bleh =P

[Mao]

for this question specifically, you should note that in , since z is raised to the odd power, its real root will take the same sign as the RHS.

[fredrick]

Cheers!

Q2)The slope field of a first order differential equation is shown below. The differential equation could be:

[IMG]http://img227.imageshack.us/img227/8088/30080603mh3.png[/img]
[IMG]http://img227.imageshack.us/img227/30080603mh3.png/1/w432.png[/img]

A)
B)
C)
D)
E)

[Mao]

technique:

pick a point on the x axis, such as x=1. at this point, the gradient is undefined (infinite).
As we move left and right (constant y, y=0), the gradient is still undefined (infinite). this is usually where there is a divide by zero, and since y is zero, we deduce that the differential equation is . that means it is either C or E

As we move up, y increases, however the gradient is negative. So we pick the negative one, C.

Taking a closer look, you might also see that tracing the slope field, you are going to get an ellipsis,
differentiating implicitly:





in this case, the ratio , so you will get an ellipsis wide as it is tall.