1. Complex numbers

Could someone, just check my answers, to see if I'm doing them right:

1) Find a real number $a$ such that:

$\mathbb{R}\left(\dfrac{1+2i}{a+3i}\right)=0$

Answer: $a=-6$

2) Factorise $x^6+1$ into real linear and quadratic factors.

$z^6+1=(z-1)(z+1)(z^2+z+1)(z^2-z+1)$

3) Find the real and imaginary parts of $(-1+i)^{77}$

Real part = $\dfrac{1}{\sqrt{2}}$
Imaginary part = $-\dfrac{1}{\sqrt{2}}$

Thanks

2. $\frac{1 + 2i}{a + 3i} = \frac{(1+2i)(a-3i)}{(a+3i)(a-3i)}$

$= \frac{a - 3i + 2ai + 6}{a^2 + 9}$

$= \frac{a + 6 + (2a - 3)i}{a^2 + 9}$

$= \frac{a + 6}{a^2 + 9} + \left(\frac{2a - 3}{a^2 + 9}\right)i$.

So the real part is $\frac{a + 6}{a^2 + 9} = 0$

$a + 6 = 0$

$a = -6$.

3. Question 1: Correct

Question 2: You can generally check identities on your own by substituting values and seeing if they hold. For example, in this question, substitute z = 1 and see!

Question 3: Incorrect
$(-1+i)^4 = ((-1+i)^2)^2 = (-2i)^2 = -4 \implies (-1+i)^{77} = (-1+i)^{76}.(-1+i) = ((-1+i)^{4})^{19}.(-1+i) = (-4)^{19}.(-1+i)$

So.....

4. Originally Posted by Isomorphism
Question 1: Correct

Question 2: You can generally check identities on your own by substituting values and seeing if they hold. For example, in this question, substitute z = 1 and see!

Question 3: Incorrect
$(-1+i)^4 = ((-1+i)^2)^2 = (-2i)^2 = -4 \implies (-1+i)^{77} = (-1+i)^{76}.(-1+i) = ((-1+i)^{4})^{19}.(-1+i) = (-4)^{19}.(-1+i)$

So.....
Wouldn't that mean for question 2 that I'm incorrect.

5. 2. $x^6 + 1 = (x^2)^3 + 1^3$

$= (x^2 + 1)(x^4 - x^2 + 1)$.

This can not be factored any further over the reals.

6. With question 3, i used the polar form of -1+i and that's how I arrived at my answer

7. Show us your solution and we will tell you where you went wrong.

8. Originally Posted by Isomorphism
Show us your solution and we will tell you where you went wrong.
Another silly error. I worked out where my mistake was. Although I prefer your method

9. $(-1 + i)^{77} = \left(\sqrt{2}\,\textrm{cis}\,\frac{3\pi}{4}\right )^{77}$

$= 2^{\frac{77}{2}}\,\textrm{cis}\,\frac{231\pi}{4}$

$= 2^{\frac{77}{2}}\,\textrm{cis}\,\frac{7\pi}{4}$

$= 2^{\frac{77}{2}}\left(\cos{\frac{7\pi}{4}} + i\sin{\frac{7\pi}{4}}\right)$

$= 2^{\frac{77}{2}}\left(\cos{\frac{\pi}{4}} - i\sin{\frac{\pi}{4}}\right)$

$= 2^{\frac{77}{2}}(2^{-\frac{1}{2}} - 2^{-\frac{1}{2}}i)$

$= 2^{38} - 2^{38}i$.

So the real part is $2^{38}$ and the imaginary part is $-2^{38}$.

10. Originally Posted by Prove It
2. $x^6 + 1 = (x^2)^3 + 1^3$

$= (x^2 + 1)(x^4 - x^2 + 1)$.

This can not be factored any further over the reals.
Ahem.

Doesn't this contradict the theorem stating that any polynomial over the reals can be factored into a product of linear and quadratic factors (over the reals)?

11. Originally Posted by awkward
Ahem.

Doesn't this contradict the theorem stating that any polynomial over the reals can be factored into a product of linear and quadratic factors (over the reals)?
I don't know what theorem you are referring to, but if you try to factorise $x^4 - x^2 + 1$...

$x^4 - x^2 + 1 = x^4 - x^2 + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + 1$

$= \left(x^2 - \frac{1}{2}\right)^2 - \frac{1}{4} + 1$

$= \left(x^2 - \frac{1}{2}\right)^2 + \frac{3}{4}$.

This does not look like it can be factorised using the Difference of Two Squares to me...

12. Originally Posted by Prove It
This does not look like it can be factorised using the Difference of Two Squares to me...
Actually,

$x^4 -x^2 + 1 = (x^4 + 2x^2 + 1) - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2 = (x^2 + 1 - \sqrt{3}x)(x^2 + 1 + \sqrt{3}x)$

Originally Posted by Prove It
I don't know what theorem you are referring to, but if you try to factorise $x^4 - x^2 + 1$...
Actually awkward is referring to the fundamental theorem of algebra (FTA) and the fact that if a complex number 'z' is a root of a polynomial 'p(x)' with real co-effs then so is the complex conjugate of z.
FTA claims there exists a root for p(x):
*If it is a real number a, then p(x) = (x-a)q(x).
* If the root is complex, say z, then so is the conjugate of z and thus the quadratic constructed using z and conjugate of z will then be a factor of p. But the quadratic with z and the conjugate of z as roots has real coefficients. So p(x) = (ax^2 + bx + c)q(x).

Notice that q(x) is a polynomial with real co-effs of strictly lesser degree. Now we can repeat the above steps on q(x) (and its factors) to break it into linear or quadratic factors again. Thus p(x) can be written as product of linear and quadratic polynomials.