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## Complex Numbers

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Algebra Manipulation

Simplify Fractions

Argand Diagrams

Conjugate Roots Theorem

Polar/Cartesian Conversion

De Moivre's Theorem

## 2022 Paper 1 Question 3

(a) $$z=6+2i$$, where $$i^2=-1$$.

(i) Show that $$z-iz=8-4i$$.

(ii) Show that $$|z|^2+|iz|^2=|z-iz|^2$$.

(iii) The circle $$c$$ passes through the points $$z$$, $$iz$$ and $$0$$, as shown in the diagram below (not to scale). $$z$$ and $$iz$$ are endpoints of a diameter of the circle.

Find the area of the circle $$c$$ in terms of $$\pi$$.

(iii) $$20\pi\mbox{ square units}$$

Solution

(i)

\begin{align}z-iz&=(6+2i)-i(6+2i)\\&=6+2i-6i-2i^2\\&=6-4i+2\\&=8-4i\end{align}

as required.

(ii)

\begin{align}iz&=i(6+2i)\\&=6i+2i^2\\&=-2+6i\end{align}

\begin{align}|z|^2+|iz|^2=|z-iz|^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(6^2+2^2)+(2^2+6^2)=8^2+4^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}36+4+4+36=64+16\end{align}

\begin{align}\downarrow\end{align}

\begin{align}80=80\end{align}

Therefore, the relation is true.

(iii)

\begin{align}r&=\frac{|z-iz|}{2}\\&=\frac{\sqrt{8^2+4^2}}{2}\\&=\frac{\sqrt{80}}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A&=\pi r^2\\&=\pi\left(\frac{\sqrt{80}}{2}\right)^2\\&=\pi\left(\frac{80}{4}\right)\\&=20\pi\mbox{ square units}\end{align}

Video Walkthrough
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(b) $$(\sqrt{3}-i)^9$$ can be written in the form $$a+ib$$, where $$a,b\in\mathbb{Z}$$ and $$i^2=-1$$.

Use de Moivre’s Theorem to find the value of $$a$$ and the value of $$b$$.

$$a=0$$ and $$b=512$$

Solution

Modulus

\begin{align}r&=\sqrt{(\sqrt{3})^2+1^2}\\&=\sqrt{3+1}\\&=\sqrt{4}\\&=2\end{align}

$\,$

Reference Angle

\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A&=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\\&=30^{\circ}\end{align}

$\,$

Argument

\begin{align}\downarrow\end{align}

\begin{align}\theta&=360^{\circ}-30^{\circ}=330^{\circ}\end{align}

$\,$

Polar Form

\begin{align}\sqrt{3}-i=2(\cos330^{\circ}+i\sin330^{\circ})\end{align}

$\,$

De Moivre’s Theorem

\begin{align}(\sqrt{3}-i)^9&=2^9[\cos9(330^{\circ})+i\sin9(330^{\circ})]\\&=512[\cos2970^{\circ}+i\sin2970^{\circ}]\\&=512(0+i)\\&=0+512i\end{align}

\begin{align}\downarrow\end{align}

$$a=0$$ and $$b=512$$

Video Walkthrough

## 2021 Paper 1 Question 1

(a) $$\dfrac{(4-2i)}{(2+4i)}=0+ki$$, where $$k\in\mathbb{Z}$$, and $$i^2=-1$$. Find the value of $$k$$.

$$k=-1$$

Solution

\begin{align}\frac{(4-2i)}{(2+4i)}&=\frac{4-2i}{2+4i}\times\frac{2-4i}{2-4i}\\&=\frac{8-16i-4i+8i^2}{4-8i+8i-16i^2}\\&=\frac{-20i}{20}\\&=-i\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=-1\end{align}

Video Walkthrough
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(b) Find $$\sqrt{-5+12i}$$.
Give both of your answers in the form $$a+bi$$, where $$a,b\in\mathbb{R}$$.

$$2+3i$$ and $$-2-3i$$

Solution

\begin{align}\sqrt{-5+12i}=a+bi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-5+12i=(a+bi)^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-5+12i=(a^2-b^2)+2abi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-5=a^2-b^2\end{align}

\begin{align}12=2ab\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a^2-b^2=-5\end{align}

\begin{align}a=\frac{6}{b}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{6}{b}\right)^2-b^2=-5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{36}{b^2}-b^2=-5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}36-b^4=-5b^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b^4-5b^2-36=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(b^2+4)(b^2-9)=0\end{align}

and therefore $$b=\pm 3$$ (since $$b\in\mathbb{R}$$). Hence

\begin{align}a&=\frac{6}{b}\\&=\frac{6}{\pm3}\\&=\pm2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sqrt{-5+12i}=2+3i\end{align}

or

\begin{align}\sqrt{-5+12i}=-2-3i\end{align}

Video Walkthrough
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(c) Use De Moivre’s theorem to find the three roots of $$z^3=-8$$.
Give each of your answers in the form $$a+bi$$, where $$a,b\in\mathbb{R}$$, and $$i^2=-1$$.

$$1+\sqrt{3}i$$, $$-2$$ and $$1-\sqrt{3}i$$

Solution

\begin{align}r=8\end{align}

and

\begin{align}\theta=\pi+2\pi n\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z^3=8[\cos(\theta+2\pi n)+i\sin(\theta+2\pi n)]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z&=8^{1/3}[\cos(\theta+2\pi n)+i\sin(\theta+2\pi n)]^{1/3}\\&=2\left[\cos\left(\frac{\pi+2\pi n}{3}\right)+i\sin\left(\frac{2\pi n}{3}\right)\right]\end{align}

$\,$

$$\mathbf{n=0}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (0)}{3}\right)+i\sin\left(\frac{\pi+2\pi (0)}{3}\right)\right]\\&=1+\sqrt{3}i\end{align}

$\,$

$$\mathbf{n=1}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (1)}{3}\right)+i\sin\left(\frac{\pi+2\pi (1)}{3}\right)\right]\\&=-2\end{align}

$\,$

$$\mathbf{n=2}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (2)}{3}\right)+i\sin\left(\frac{\pi+2\pi (2)}{3}\right)\right]\\&=1-\sqrt{3}i\end{align}

Video Walkthrough

## 2020 Paper 1 Question 2

(a) Find the two complex numbers $$z_1$$ and $$z_2$$ that satisfy the following simultaneous equations, where $$i^2=-1$$:

\begin{align}iz_1&=-4+3i\\3z_1-z_2&=11+17i\end{align}

Write your answers in the form $$a+bi$$ where $$a,b\in\mathbb{Z}$$.

$$z_1=3+4i$$ and $$z_2=-2-5i$$

Solution

\begin{align}iz_1=-4+3i\end{align}

\begin{align}\downarrow\end{align}

\begin{align}i(iz_1)=i(-4+3i)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-z_1=-4i+3i^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z_1=3+4i\end{align}

and

\begin{align}3(3+4i)-z_2=11+17i\end{align}

\begin{align}\downarrow\end{align}

\begin{align}9+12i-z_2=11+17i\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z_2&=-11-17i+9+12i\\&=-2-5i\end{align}

Video Walkthrough
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(b) The complex numbers $$3+2i$$ and $$5-i$$ are the first two terms of a geometric sequence.

(i) Find $$r$$, the common ratio of the sequence.

Write your answer in the form $$a+bi$$ where $$a,b,\in\mathbb{Z}$$.

(ii) Use de Moivre’s Theorem to find $$T_9$$, the ninth term of the sequence.
Write your answer in the form $$a+bi$$ where $$a,b,\in\mathbb{Z}$$.

(i) $$r=1-i$$

(ii) $$48+32i$$

Solution

(i)

\begin{align}r&=\frac{T_2}{T_1}\\&=\frac{5-i}{3+2i}\\&=\frac{5-i}{3+2i}\times\frac{3-2i}{3-2i}\\&=\frac{15-10i-3i+2i^2}{9-6i+6i-4i^2}\\&=\frac{13-13i}{13}\\&=1-i\end{align}

(ii)

\begin{align}T_n=ar^{n-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_9&=ar^8\\&=(3+2i)(1-i)^8\end{align}

For $$(1-i)$$, the modulus is

\begin{align}r&=\sqrt{1^2+(-1)^2}\\&=\sqrt{2}\end{align}

and, since it is in the fourth quadrant, the argument is

\begin{align}\theta&=2\pi-\tan^{-1}\left(\frac{1}{1}\right)\\&=2\pi-\frac{\pi}{4}\\&=\frac{7\pi}{4}\end{align}

Therefore, we have

\begin{align}T_9&=(3+2i)(1-i)^8\\&=(3+2i)\left[\sqrt{2}\left[\cos\left(\frac{7\pi}{4}\right)+i\sin\left(\frac{7\pi}{4}\right)\right]\right]^8\\&=(3+2i)(\sqrt{2})^8\left[\cos\left(\frac{7\pi(8)}{4}\right)+i\sin\left(\frac{7\pi(8)}{4}\right)\right]\\&=(3+2i)(16)(\cos14\pi+i\sin14\pi)\\&=(3+2i)(16)(1+0i)\\&=48+32i\end{align}

Video Walkthrough

## 2019 Paper 1 Question 5

(a) $$3+2i$$ is a root of $$z^2+pz+q=0$$, where $$p,q\in\mathbb{R}$$, and $$i^2=-1$$.
Find the value of $$p$$ and the value of $$q$$.

$$p=-6$$ and $$q=13$$

Solution

\begin{align}(3+2i)^2+p(3+2i)+q=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}9+12i-4+3p+2pi+q=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5+3p+q)+(12+2p)i=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5+3p+q)+(12+2p)i=0\end{align}

\begin{align}\downarrow\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}5+3p+q=0\end{align}

\begin{align}12+2p=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5+3p+q=0\end{align}

\begin{align}p=-6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5+3(-6)+q=0\end{align}

\begin{align}p=-6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q=13\end{align}

\begin{align}p=-6\end{align}

Video Walkthrough
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(b)

(i) $$v=2-2\sqrt{3}i$$. Write $$v$$ in the form $$r(\cos\theta+i\sin\theta)$$, where $$r\in\mathbb{R}$$ and $$0\leq\theta\leq2\pi$$.

(ii) Use your answer to part (b)(i) to find the two possible values of $$w$$, where $$w^2=v$$.
Give your answers in the form $$a+ib$$, where $$a,b\in\mathbb{R}$$.

(i) $$v=4\left[\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right]$$

(ii) $$w=-\sqrt{3}+i$$ and $$w=\sqrt{3}-i$$

Solution

(i)

Modulus

\begin{align}r&=\sqrt{2^2+(-2\sqrt{3})^2}\\&=\sqrt{4+12}\\&=\sqrt{16}\\&=4\end{align}

$\,$

Argument

Reference Angle:

\begin{align}A&=\tan^{-1}\left(\frac{2\sqrt{3}}{2}\right)\\&=\frac{\pi}{3}^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=2\pi-\frac{\pi}{3}\\&=\frac{\pi}{3}\end{align}

$\,$

Polar Form

\begin{align}v=4\left[\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right]\end{align}

(ii)

\begin{align}w&=\pm\left[4\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right)\right]^{1/2}\\&=\pm2\left[\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right]\\&=\pm(-\sqrt{3}+i)\end{align}

\begin{align}\downarrow\end{align}

$$w=-\sqrt{3}+i$$ and $$w=\sqrt{3}-i$$

Video Walkthrough

## 2018 Paper 1 Question 4

(a) Prove, using induction, that if $$n$$ is a positive integer then

$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$, where $$i^2=-1$$.

Solution

For $$n=1$$, the formula states that

\begin{align}(\cos\theta+i\sin\theta)^1=\cos(1\theta)+i\sin(1\theta)\end{align}

\begin{align}\cos\theta+i\sin\theta=\cos\theta+i\sin\theta\end{align}

and therefore it is true for $$n=1$$.

Hence, let us assume that it is true for $$n=k$$, i.e. that

\begin{align}(\cos\theta+i\sin\theta)^k=\cos(k\theta)+i\sin(k\theta)\end{align}

Using this assumption, we now wish to prove that it is also true for $$n=k+1$$, i.e. that

\begin{align}(\cos\theta+i\sin\theta)^{k+1}=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}

Let us look at the left hand side:

\begin{align}(\cos\theta+i\sin\theta)^{k+1}&=(\cos\theta+i\sin\theta)^{k}(\cos\theta+i\sin\theta)^{1}\\&=[\cos(k\theta)+i\sin(k\theta)](\cos\theta+i\sin\theta)\theta]\\&=[\cos(k\theta)-\sin(k\theta)\sin\theta]+i[\cos(k\theta)\sin\theta+\cos\theta\sin(k\theta)]\\&=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}

which is indeed the right hand side.

Therefore, the formula is true for all positive integers $$n$$.

Video Walkthrough
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(b) Hence, or otherwise, find $$\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)^3$$ in its simplest form.

$$1$$

Solution

Modulus

\begin{align}r&=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\&=\sqrt{\frac{1}{4}+\frac{3}{4}}\\&=\sqrt{1}\\&=1\end{align}

$\,$

Argument

\begin{align}\tan\theta&=-\frac{\sqrt{3}/2}{1/2}\\&=-\sqrt{3}\end{align}

Reference Angle:

\begin{align}\tan A=\sqrt{3}\end{align}

\begin{align}\downarrow\end{align}

As $$\theta$$ is in the second quadrant:

\begin{align}\theta&=\pi-A\\&=\pi-\frac{\pi}{3}\\&=\frac{2\pi}{3}\end{align}

$\,$

Polar Form

\begin{align}-\frac{1}{2}+\frac{\sqrt{3}}{2}i&=1\left[\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right]\\&=\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^3&=\left[\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right]^3\\&=\cos(3)\left(\frac{2\pi}{3}\right)+i\sin(3)\left(\frac{2\pi}{3}\right)\\&=\cos(2\pi)+i\sin(2\pi)\\&=1+0i\\&=1\end{align}

Video Walkthrough

## 2017 Paper 1 Question 2

$$z=-\sqrt{3}+i$$, where $$i^2=-1$$.

(a) Use De Moivre’s Theorem to write $$z^4$$ in the form ܽ$$a+b\sqrt{c}i$$, where ܽ$$a$$, $$b$$, and $$c\in\mathbb{Z}$$.

$$z^4=-8-8\sqrt{3}i$$

Solution

Modulus

\begin{align}r&=\sqrt{(-\sqrt{3})^2+1^2}\\&=\sqrt{3+1}\\&=2\end{align}

$\,$

Argument

\begin{align}\tan\theta=-\frac{1}{\sqrt{3}}\end{align}

Reference Angle:

\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=\frac{\pi}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\pi-\frac{\pi}{6}\\&=\frac{5\pi}{6}\end{align}

$\,$

Polar Form

\begin{align}z=2(\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z^4&=2^4(\cos\left(\frac{5(4)\pi}{6}\right)+i\sin\left(\frac{5(4)\pi}{6}\right)\\&=-8-8\sqrt{3}i\end{align}

Video Walkthrough
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(b) The complex number $$w$$ is such that $$|w|=3$$ and $$w$$ makes an angle of $$30^{\circ}$$ with the positive sense of the real axis. If $$t=zw$$, write $$t$$ in its simplest form.

$$t=-6$$

Solution

\begin{align}w=3\left[\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=zw\\&=2\left[\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)\right]\times3\left[\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right]\\&=6(\cos\pi+i\sin\pi)\\&=6(-1+0i)\\&=-6\end{align}

Video Walkthrough

## 2016 Paper 1 Question 1

(a) $$(-4+3i)$$ is one root of the equation $$az^2+bz+c=0$$, where $$a,b,c\in\mathbb{R}$$, and $$i^2=-1$$.
Write the other root.

$$-4-3i$$

Solution

$$-4-3i$$

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Use De Moivre’s Theorem to express $$(1+i)^8$$ in its simplest form.

$$16$$

Solution

\begin{align}z=1+i\end{align}

$\,$

Modulus

\begin{align}r&=\sqrt{1^2+1^2}\\&=\sqrt{2}\end{align}

$\,$

Argument

\begin{align}\tan\theta&=\frac{1}{1}\\&=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\tan^{-1}(1)\\&=\frac{\pi}{4}\end{align}

$\,$

Polar Form

\begin{align}(i+1)=\sqrt{2}\left[\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(i+1)^8&=\sqrt{2}^8\left[\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right]^8\\&=16\left[\cos\left(\frac{8\pi}{4}\right)+i\sin\left(8\frac{\pi}{4}\right)\right]\\&=16[\cos(2\pi)+i\sin(2\pi)]\\&=16(1=)i)\\&=16\end{align}

Video Walkthrough
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(c) $$(1+i)$$ is a root of the equation $$z^2+(-2+i)z+3-i=0$$.
Find its other root in the form $$m+ni$$, where $$m,n\in\mathbb{R}$$, and $$i^2=-1$$.

$$1-2i$$

Solution

\begin{align}z&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2+i)\pm\sqrt{(-2+i)^2-4(1)(3-i)}}{2(1)}\\&=\frac{2-i\pm\sqrt{4-4i+i^2-12+4i}}{2}\\&=\frac{2-i\pm\sqrt{-9}}{2}\\&=\frac{2-i\pm3i}{2}\end{align}

\begin{align}\downarrow\end{align}

$$z=1-2i$$ or $$z=1+i$$

Therefore, the other root is $$z=1-2i$$.

Video Walkthrough

## 2015 Paper 1 Question 4

(a) The complex numbers $$z_1$$, $$z_2$$ and $$z_3$$ are such that $$\dfrac{2}{z_1}=\dfrac{1}{z_2}+\dfrac{1}{z_3}$$, $$z_2=2+3i$$ and $$z_3=3-2i$$, where $$i^2=-1$$. Write $$z_1$$ in the form $$a+bi$$, where $$a,b\in\mathbb{Z}$$.

$$5+i$$

Solution

\begin{align}\frac{2}{z_1}&=\frac{1}{2+3i}+\frac{1}{3-2i}\\&=\frac{3-2i+2+3i}{(2+3i)(3-2i)}\\&=\frac{5+i}{6-4i+9i-6i^2}\\&=\frac{5+i}{12+5i}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z_1&=\frac{24+10i}{5+i}\\&=\frac{24+10i}{5+i}\times\frac{5-i}{5-i}\\&=\frac{120-24i+50i-10i^2}{25-5i+5i-i^2}\\&=\frac{130+26i}{26}\\&=5+i\end{align}

Video Walkthrough
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(b) Let $$\omega$$ be a complex number such that $$w^n=1$$, $$\omega\neq 1$$ and $$S=1+\omega+\omega^2+…+\omega^{n-1}$$. Use the formula for the sum of a finite geometric series to write the value of $$S$$ in its simplest form.

$$S=0$$

Solution

\begin{align}a=1&&r=\omega\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S&=\frac{a(1-r^n)}{1-r}\\&=\frac{1(1-\omega^n)}{1-\omega}\\&=\frac{1(1-1)}{1-\omega}\\&=0\end{align}

Video Walkthrough