---

title: Calculas Scholarship T1W7 (Complex Algebra)
tags: Calculus Scholarship
description: Term 1 Week 7 Complex Algebra
type: pdf
slideOptions:
theme: white

---

###### tags: `Calculus Scholarship`

### **Q1 (for L2 students only)**

I will just show one example here, the rest should follows

> Proof:
let $z = x + iy$, then $\bar{z} = x - iy$, we have $$LHS = z\bar{z} = (x+iy)(x-iy)= x^2 + ixy - ixy -i^2y^2 = x^2 + y^2$$
$$RHS = |z|^2 = (\sqrt{(x^2+y^2)})^2 = x^2 + y^2$$
So LHS = RHS, $\blacksquare$

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### **Q2**

The key points have been mentioned in the **Hint**.

> Solve:
As (2+3i) is one root, we have (2-3i) is another root, so by sum and product rules, assuming a = 1, we have b = -4 and c = (2+3i)(2-3i) = 13. Thus, $x^2 - 4x + 13$ is a factor of P(z).

> P(-1) = 0, so (x+1) is a factor of P(z), with long division, we have $P(z) = (x+1)(x^2 - 4x + 13)$. So the three roots are 1, 2+3i and 2-3i

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### **Q3**

> Solve:
$z + \frac{1}{\bar{z}} = \bar{z} + \frac{1}{z} \implies \frac{\bar{z}z+1}{\bar{z}} = \frac{\bar{z}z+1}{z}\implies z=\bar{z}$
So we must have y = -y, which is y = 0. Also x can be any real number except for 0, as $\frac{1}{z} and \frac{1}{\bar{z}}$ would be undefinded if $z = 0$

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### **Q4**

Please refers to

> Group 1: **Andrew** or **Suyoung**'s answers
> Group 2:
> Another way to think of Algebraicly is:
$$\bar{z} = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$$, so $\bar{z}^2 = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then $(\bar{z}^2)^2 = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$

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### **Q5**

(a). omitted, as it is simply calculation of $z^2$

(b). <Also, read **Andrew**'s answer>

What we need to do is **following the definition and examples**

- for $c = 1+\frac{i}{2}$

$|f(0)| = |c| = \sqrt{1 + \frac{1}{2}^2} = \frac{\sqrt{5}}{2} < 2$

$|f^2(0)| = |c^2 + c| = |(1+\frac{i}{2})^2 + (1+\frac{i}{2})| = |\frac{7}{4} + \frac{3}{2}i| = \frac{\sqrt{65}}{4} > 2$, so $1+\frac{i}{2}$ is not in the Mandelbrot set.

- for $c = i$

$|f(0)| = |c| = \sqrt{1^2} = 1 < 2$

$|f^2(0)| = |c^2 + c| = |i^2 + i| = |-1+i| = \sqrt{2} < 2$,

$|f^3(0)| = |c^2 + c| = |(-1+i)^2 + i| = |-2i+i| = |-i| = 1 < 2$,

$|f^4(0)| = |c^2 + c| = |(-i)^2 + i| = |-1+i| = \sqrt{2} < 2$,

As we can see the periodic feature from the function and its modular from $|f^2(0)|$ to $|f^4(0)|$, we can conclude that i belongs to the Mandelbrot set.