Hello,
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!
I'll provide hints for you and then a full solution to some of the questions in the spoiler!
Q4Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\]
Q4 - full solution
As hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]
We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]
Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)
So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]
Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]
Q5Q5 - full solution
Let:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]
We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]
Q6Hint 1: You may like to take a geometric approach
Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]
Q6 - full solution
We note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]
By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]
We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]
Q8Hint: Best bet would be to prove it by induction.
Q8 - full solution
Base case is just the simple triangle inequality.
When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]
When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.
Q9Q9 - full solution
If the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]
and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]