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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... |
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Algebra Manipulation |
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Simplify Fractions |
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Argand Diagrams |
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Conjugate Roots Theorem |
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Polar/Cartesian Conversion |
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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 \(\pi\).
(i) The answer is already in the question!
(ii) The answer is already in the question!
(iii) \(20\pi\mbox{ square units}\)
(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}
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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 \(b=512\)
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
Fourth Quadrant
\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 \(b=512\)
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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 \(k\).
\(k=-1\)
\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}
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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 \(-2-3i\)
\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}
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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 \(1-\sqrt{3}i\)
\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}
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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 \(z_2=-2-5i\)
\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}
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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) \(48+32i\)
(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}
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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 \(q=13\)
\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}
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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 \(w=\sqrt{3}-i\)
(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 \(w=\sqrt{3}-i\)
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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 \(i^2=-1\).
The answer is already in the question!
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 \(n\).
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(b) Hence, or otherwise, find \(\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)^3\) in its simplest form.
\(1\)
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}
\begin{align}A&=\frac{\pi}{3}\mbox{ rad}\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}
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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 \(c\in\mathbb{Z}\).
\(z^4=-8-8\sqrt{3}i\)
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}
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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.
\(t=-6\)
\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}
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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.
\(-4-3i\)
\(-4-3i\)
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(b) Use De Moivre’s Theorem to express \((1+i)^8\) in its simplest form.
\(16\)
\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}
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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 \(i^2=-1\).
\(1-2i\)
\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 \(z=1-2i\).
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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 \(a,b\in\mathbb{Z}\).
\(5+i\)
\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}
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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.
\(S=0\)
\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}
