Complex number problem

[Ggdivhjkjl]

Greetings, I'm not likely to post again but I was having trouble with some maths and so I ran a search and found these forums in the hope somebody could answer my question.

 

I understand this:

 

"

Given z = root(-9) (note: that's root of minus nine.)

then z2 = (-9) .....................(1) (note: that's z squared)

Let z = r cis o..................(2) (where o is phi)

then z2 = r2 cis 2o

Also, -9 = 9 cis p (where p is pi)

 

equation (1) can be rewritten as:

r2 cis 2o = 9 cis p

r2 (cos 2o + isin 2o) = 9 (cos p+ isin p)

hence, r2 = 9

 

equating the terms inside the brackets gives:

2o = p + 2kp (where k is an integer to account for coterminal angel solutions)

 

Solving:

r =root(9) = 3

o= (p/2) + (2kp/2) = (p/2) + kp (where k is an integer)

When k = 0, o = (p/2)

When k = 1, o = (p/2) + 1(p) = (3p/2)

which is coterminal with (-p/2)

 

Substituting these results in (2) gives:

 

z = r cis o

z = 3 cis (p/2) or z = 3 cis (-p/2)

hence z = 3i or -3i

"

 

How then do I find all the solutions for this:

 

"

Given z(z+5)=176 find all solutions for z.

 

The real ones are easy:

 

x(x+5) = 176

x2 + 5x = 176

x2 + 5x - 176 = 0

Hence x = {-5 plus or minus root[25-4(1)(-176)]/2(1)}

= 11 or -16

 

Are there any imaginary solutions or was the wording of the question just a trick to get us thinking?

 

If there are any could somebody please show me how to find them?

 

Thank you all.

[Ggdivhjkjl]

I probably shouldn't have posted this under Calculus should I?

[Dave]

The equation you gave is a quadratic, which can only possibly have two solutions at most. If both solutions are real, then there are no imaginary roots to the equation.

 

You can check it out by plotting the graph if you really wanted to - if the graph doesn't cut the x-axis then there are no real roots to the equation.

[Ggdivhjkjl]

Sorry for taking so long to reply but thank you nonetheless.

I had not noticed that at the time I posted my question obviously and I am often easily puzzled when asked to find "all solutions" as I expect there to be imaginary ones in a question such as this.

Thank you again and farewell.

[Dave]

Not a problem

[EvoN1020v]

This thread should be posted in the "Trigonometry" section, as I have learnt those [math]z = rcis\phi[/math] last semester.