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If imaginary part = 0, then number is real (a real number is still a complex number), so you should be OK. 7 = 7 + 0i (but we don't usually write it like this)

Over the last 40 years or so, geometry has not received as much attention as they used to. Many questions on complex numbers are better approached by geometry (easier, shorter, visually-clearer, neater solutions). So if you want to do well in complex numbers, make sure you have a good foundation...

\text {If you draw the diagram, you can show triangle AOC and triangle COB are similar. Why?}\\ \\ \frac {1}{z} = \frac {1\times \bar z}{z \times \bar z} = \frac {\bar z}{|z|^2} \\ \\ \text { and } \bar z \text{ is a reflection of z on the x-axis so that } \angle AOC = \angle COB \\ \\ \text {...

When we talk about Selective Schools, we usually have the public high schools in mind. Their teachers are paid the standard public school rates. Sydney Grammar is selective, but is a PRIVATE school. So can afford to pay its teachers more. I don't know anything about double pay. I've never been...

\therefore z^2(5z^2-11z^1 + 16 -11z^{-1} +5z^{-2})=0\\ \\ \therefore (5(z^2+z^{-2}) -11(z^1+z^{-1}) + 16 )= 0 $ if $ z \neq 0\\ \\ z=cis \theta \implies z^n = cis (n\theta )$ and $ z^{-n} = cis( {-n\theta} )\implies z^n+z^{-n} = 2cos(n\theta)\\ \\ \therefore 10cos(2\theta) -22cos(\theta )+ 16 =...

Q18 \text {Given: } z = 2cis\theta \\ \\ \therefore \frac{1}{1-z} =\frac {\overline {1-z}}{(1-z)(\overline{1-z})}=\frac {1-\bar z}{(1-z)(1-\bar z)} = \frac {1-\bar z}{1-z-\bar z + z\bar z}\\ \\ = \frac {1 -2cos \theta + 2i sin \theta}{1-4cos \theta + 4} \\ \\ = \frac...

OK. Sorry, I have not spelt out in more detail because I really struggle to LaTeX my solution; so have to skip intermediate steps. Which part would you like me to explain more fully?

(a) z^{2n} - 1 = (z-1)(1+z+z^2 + \cdots + z^{2n-1})\\ \\ \therefore 1+z+z^2 + \cdots + z^{2n-1} = \frac{z^{2n}-1}{z-1} (z \neq 1))\\ \\ = \frac{(cis{\frac{\pi}{n}})^{2n}-1}{cis{\frac {\pi}{n}}-1} = \frac{cis{\frac {2n\pi}{n}}-1}{\cdots} = \frac {cis{2\pi}-1}{\cdots} = \frac{0}{\cdots} = 0 (b)...

That's essentially what I said before. The simplest explanation.

This seems a sound approach.

Thank you very much for pointing out my typo. Now corrected.

a) r = \sqrt 2 \\ \\ (1+i)^n + (1-i)^n = (rcis\frac {\pi}{4})^n +(rcis{\frac{-\pi}{4}})^n \\ \\ = r^n cis \frac{n\pi}{4}) + r^n cis \frac{-n\pi}{4}\\ \\ =r^n(cos \frac {n\pi}{4} + isin \frac {n\pi}{4}) + r^n(cos \frac{n\pi}{4} - isin \frac{n\pi}{4})\\ \\ = 2r^n cos \frac {n\pi}{4}\\ \ = real...

9(9^{k+1} - 1) - 8(k+1) = 9([9(9^k - 1) - 8k + 8 + 8k]) - 8(k+1) \\ \\= 9(9(9^k - 1) - 8k )+72k + 72 - 8k - 8 \\ \\ = 9 \times 64M +64k + 64 = 64 \times (9M + k + 1) Essentially the same as Trebla's

1st one is very legible and above average by today's low standards. No need to improve. 2nd one is terrible.

I don't know what SLR is all about. Bloody do Maths Ext 1 and if necessary, drop SLR!

Then P(x) will always be positive, i.e. cannot become zero. That means it cannot have real roots. That is you cannot find a real number "a" say, such that P(a) = 0. To expand: x^4 >= 0; Ax^2 >= 0; B > 0 ==> P(x) > 0 So the curve of y = P(x) cannot cross the x-axis; so no real roots !!!!!

At least Maths Advanced; preferably Maths Ext 1. It is not as if you are not capable of handling Advanced, like the many who choose Standard. Although the maths required to understand HSC Physics is not very difficult, a background in calculus would help you greatly. You need to be sufficiently...

The diagonals of a parallelogram bisect each other. So in a //gram ABCD, the diagonals AC and BD bisect each other. What this means is that AC cuts BD into 2 equal halves and BD cuts AC into 2 equal halves. In an arbitrary quadrilateral ABCD, diagonal AC may cut diagonal BD into 2 equal halves...

B

C) & D) can be discounted immediately because the don't have modulus (length) 1.