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Complex numbers questions

Two questions here. The 1st I've done and need clarification on whether it's the right method. The other I have no idea.

Show, by constructing an example, that non-real numbers b,c may be found so that the quadratic equation x2+bx+c=0x^2 + bx + c = 0 has a real root. Is it possible for this equation to have two real roots if b,c are non-real?

Ok, for the first part I'm guessing that if they are asking for a real root then the discriminant has to be equal to zero (because you'll get one, repeated root). I'm also assuming that by non-real they mean b and c have to be either complex or imaginary (since all other number systems are contained in the real number system).

b2−4ac=0b^2 - 4ac = 0

since a = 1

b2−4c=0b^2 - 4c = 0

now let b=4i,c=i b = \sqrt{4i}, c = i

(4i)2−4i=0(\sqrt{4i})^2 - 4i = 0

and this equates to zero (obviously). For the second part the discriminant needs to be greater than zero, i.e.

b^2 - 4ac > 0

I THINK that it is possible for the equation to have two real roots if b,c are non-real, the condition being b>c. My only concern being if the complex number system follows the same rules as the other number systems (i.e. when is one complex number bigger than another? When the real part of one complex number is bigger than the other? Or the imaginary part of one complex number is bigger than the other? Or both?).

Now, to my second question

Prove that 1+sin⁡θ+icos⁡θ1+sin⁡θ−icos⁡θ=sin⁡θ+icos⁡θ\frac{1 + \sin \theta + i\cos \theta}{1 + \sin \theta - i\cos \theta} = \sin \theta + i\cos \theta

Any hints on how to start this?
Reply 1
Avatar for nota bene
nota bene
15
For the second question I'd guess you can start with multiplying the lhs with the conjugate of the denominator.
Reply 2
Avatar for Syres4me
Syres4me
OP
2
Hmm.....why didn't I think of that? I started by multiplying both top and bottom by the denominator initially.
Reply 3
Avatar for Morbo
Morbo
17
Syres4me
when is one complex number bigger than another? When the real part of one complex number is bigger than the other? Or the imaginary part of one complex number is bigger than the other? Or both?

It makes no sense to talk about one complex number being bigger than another. You can compare the size of the modulus or argument of the numbers (in the same way you can compare the size of the real or imaginary components), but that is all.
Reply 4
Avatar for Zhen Lin
Zhen Lin
12
For your first question, bear in mind the following properties of roots of quadratics. Let A, B be roots of ax2+bx+c=0ax^2 + bx + c = 0. Then, -b/a = A + B, and c/a = AB. From the first property you can get answer to the first half of your first question, and from the second property you can get an answer to the second half.
Reply 5
Avatar for DFranklin
DFranklin
18
Syres4me
Two questions here. The 1st I've done and need clarification on whether it's the right method. The other I have no idea.

Show, by constructing an example, that non-real numbers b,c may be found so that the quadratic equation x2+bx+c=0x^2 + bx + c = 0 has a real root. Is it possible for this equation to have two real roots if b,c are non-real?

Ok, for the first part I'm guessing that if they are asking for a real root then the discriminant has to be equal to zero (because you'll get one, repeated root). I'm also assuming that by non-real they mean b and c have to be either complex or imaginary (since all other number systems are contained in the real number system).

b2−4ac=0b^2 - 4ac = 0

since a = 1

b2−4c=0b^2 - 4c = 0

now let b=4i,c=i b = \sqrt{4i}, c = i

(4i)2−4i=0(\sqrt{4i})^2 - 4i = 0

and this equates to zero (obviously).True. But unfortunately b isn't real so −b±b2−4ac-b \pm \sqrt{b^2-4ac} isn't either. In fact, you can't find a solution if the discriminant is zero for the same reason as:

For the second part the discriminant needs to be greater than zero, i.e.

b^2 - 4ac > 0

I THINK that it is possible for the equation to have two real roots if b,c are non-real, the condition being b>c.
As others have said, b > c has no meaning if b and c are complex. But in any event, your reasoning here is spurious, because if b isn't real, it doesn't help you to show that b2−4ac\sqrt{b^2-4ac} is.

The best plan here is to work the other way. Suppose α,β\alpha, \beta are roots of x2+bx+c=0x^2+bx+c = 0. Factorise, equate coefficients, and you'll get expressions for b and c in terms of α,β\alpha, \beta. So if α\alpha and β\beta are real, what can we say about b and c?
For the simplification of cosine and sine fraction, express the cosine and sine in exponential form using e^iθ =cosθ +isinθ
Reply 7
Avatar for nota bene
nota bene
15
username62978
For the simplification of cosine and sine fraction, express the cosine and sine in exponential form using e^iθ =cosθ +isinθ

That was my initial thought as well, but because I didn't spot

Spoiler



Now, having spotted that it takes me three lines of working to get the answer, so it is indeed a better method than looking at the conjugate:smile:.
(edited 4 years ago)
Reply 8
Avatar for Syres4me
Syres4me
OP
2
@ DFranklin

I understand what your saying, but why are you singling out b (inparticular) as being non-real?
Reply 9
Avatar for nota bene
nota bene
15
Both b and c are non-real by definition, read your question again.
Syres4me
@ DFranklin

I understand what your saying, but why are you singling out b (inparticular) as being non-real?
Because you are basing your argument on making b2−4ac\sqrt{b^2-4ac} real, when the quadratic formula gives the roots as −b±b2−4ac2\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2}.

In other words, to know what the roots are, you need to know both bb and b2−4ac\sqrt{b^2-4ac}, but if you know these, it doesn't matter what the value of cc is.

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