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Complex Numbers
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(BLUE = Paper 1, RED = Paper 2)
2022 Paper 1 Question 1
The complex number \(z_1\) is shown on the Argand diagram below.
(a) Using the Argand diagram:
(i) write down the values of \(z_1\) and \(\bar{z_1}\), where \(\bar{z_1}\) is the complex conjugate of \(z_1\).
(ii) plot and label \(\bar{z_1}\) on the Argand diagram above.
(i) \(z_1=-2-3i\) and \(\bar{z_1}=-2+3i\)
(ii)
(i)
\begin{align}z_1=-2-3i && \bar{z_1}=-2+3i\end{align}
(ii)
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\(z_2\) and \(z_3\) are two other complex numbers.
\(z_2=-5+3i\) and \(z_3=4-2i\), where \(i^2=-1\).
(b) Plot and label \(z_2\) and \(z_3\) on the Argand diagram from part (a).
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(c) Write \(z_2-z_3\) in the form \(a+bi\), where \(a,b,\in\mathbb{R}\), \(i^2=-1\), and hence find \(|z_2-z_3|\).
\(z_2-z_3=-9+5i\) and \(|z_2-z_3|=\sqrt{106}\)
\begin{align}z_2-z_3&=(-5+3i)-(4-2i)\\&=-5+3i-4+2i\\&=-9+5i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|z_2-z_3|&=|-9+5i|\\&=\sqrt{(-9)^2+5^2}\\&=\sqrt{106}\end{align}
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(d) Investigate if \(z_3=4-2i\) is a solution of the equation \(z^2+2iz-7i=0\).
\(z_3\) is not a solution.
\begin{align}z^2+2iz-7i&=(4-2i)^2+2i(4-2i)-7i\\&=16-16i-4+8i+4-7i\\&=16-15i\\&\neq0\end{align}
Therefore, \(z_3\) is not a solution.
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2021 Paper 1 Question 2
\(z_1=-3+4i\) and \(z_2=4+3i\), where \(i^2=-1\).
(a) Plot and label \(z_1\), \(z_2\) and \(z_1+z_2\) on the Argand Diagram.
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(b) \(z_3=\dfrac{z_1}{z_2}\). Find \(z_3\) in the form \(a+bi\), where \(a,b\in\mathbb{Z}\).
\(z_3=i\)
\begin{align}z_3&=\frac{z_1}{z_2}\\&=\frac{-3+4i}{4+3i}\\&=\frac{-3+4i}{4+3i}\times\frac{4-3i}{4-3i}\\&=\frac{-12+9i+16i+12}{16-12i+12i+9}\\&=\frac{25i}{25}\\&=i\end{align}
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(c) Find \(|\bar{z_1}-z_2|\), where \(\bar z_1\) is the complex conjugate of \(z_1\).
Give your answer in the form \(p\sqrt{q}\), where \(p\) and \(q\in\mathbb{N}\)
\(7\sqrt{2}\)
\begin{align}|\bar{z_1}-z_2|&=|(-3-4i)-(4+3i)|\\&=|-7-7i|\\&=\sqrt{(-7)^2+(-7)^2}\\&=\sqrt{49\times2}\\&=\sqrt{49}\sqrt{2}\\&=7\sqrt{2}\end{align}
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2020 Paper 1 Question 3
\(z_1=3-4i\), \(z_2=-2+i\) and \(z_3=2iz_2\), where \(i^2=-1\).
(a)
(i) Write \(z_3\) in the form \(a+bi\), where \(a,b\in\mathbb{Z}\).
(ii) Plot \(z_1\), \(z_2\) and \(z_3\) on the given Argand Diagram.
Label each point clearly.
(iii) Find \(|z_1|\).
(i) \(-2-4i\)
(ii)
(iii) \(5\)
(i)
\begin{align}z_3&=2iz_2\\&=2i(-2+i)\\&=-4i+2i^2\\&=-2-4i\end{align}
(ii)
(iii)
\begin{align}|z_1|&=\sqrt{3^2+(-4)^2}\\&=\sqrt{9+16}\\&=\sqrt{25}\\&=5\end{align}
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(b) If \(z_1\times z_4=29+3i\), write \(z_4\) in the form \(a+bi\), where \(a,b\in\mathbb{R}\).
\(3+5i\)
\begin{align}z_4&=\frac{29+3i}{z_1}\\&=\frac{29+3i}{3-4i}\\&=\frac{29+3i}{3-4i}\times\frac{3+4i}{3+4i}\\&=\frac{87+116i+9i-12}{9-12i+12i+16}\\&=\frac{75+125i}{25}\\&=3+5i\end{align}
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2019 Paper 1 Question 2
(a) The complex number \(z_1=2+i\), where \(i^2=-1\), is shown on the Argand Diagram below.
(i) \(z_2=2z_1\).
Find the value of \(z_2\), and plot and label it on the Argand Diagram.
(ii) \(\bar{z_1}\) is the complex conjugate of \(z_1\).
Write down the value of \(\bar{z_1}\), and plot and label it on the Argand Diagram.
(iii) Investigate if \(|z_2|=|z_1+\bar{z_1}|\).
(i) \(z_2=4+2i\)
(ii) \(\bar{z_1}=2-i\)
(iii) They are not equal.
(i)
\begin{align}z_2&=2z_1\\&=2(2+i)\\&=4+2i\end{align}
(ii)
\begin{align}\bar{z_1}=2-i\end{align}
(iii)
\begin{align}|\bar{z_2}|&=\sqrt{4^2+2^2}\\&=\sqrt{20}\end{align}
and
\begin{align}|z_1+\bar{z_1}|&=|(2+i)+(2-i)|\\&=|4|\\&=4\end{align}
Therefore, they are not equal.
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(b) Show that \(z_1=2+i\) is a solution of the equation \(z^2-4z+4=0\).
The answer is already in the question!
\begin{align}(2+i)^2-4(2+i)+5&=4+4i-1-8-4i+5\\&=0\end{align}
as required.
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2018 Paper 1 Question 2
\(z_1=-2+3i\) and \(z_2=-3-2i\), where \(i^2=-1\).
\(z_3=z_1-z_2\).
(a) Plot \(z_1\), \(z_2\), and \(z_3\) on the Argand Diagram.
Label each point clearly.
\begin{align}z_3&=z_1-z_2\\&=(-2+3i)-(-3-2i)\\&=-2+3i+3+2i\\&=1+5i\end{align}
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(b) Investigate if \(|z_3|=|z_1|+|z_2\).
\(|z_3|\neq|z_1|+|z_2|\)
\begin{align}|z_3|&=\sqrt{1^2+5^2}\\&=\sqrt{26}\end{align}
and
\begin{align}|z_1|+|z_2|&=\sqrt{(-2)^2+3^2}+\sqrt{(-3)^2+(-2)^2}\\&=\sqrt{13}+\sqrt{13}\\&=2\sqrt{13}\end{align}
Therefore, \(|z_3|\neq|z_1|+|z_2|\).
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(c) \(z_4=\dfrac{z_1}{z_2}\). Write \(z_4\) in the form \(x+yi\),݅ where \(x,y,\in\mathbb{R}\).
\begin{align}z_4=-i\end{align}
\begin{align}z_4&=\frac{-2+3i}{-3-2i}\\&=\frac{-2+3i}{-3-2i}\times\frac{-3+2i}{-3+2i}\\&=\frac{6-4i-9i-6}{9-6i+6i+4}\\&=\frac{-13i}{13}\\&=-i\end{align}
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2017 Paper 1 Question 2
(a) The complex number \(z_1=a+bi\), where \(i^2=-1\), is shown on the Argand Diagram below.
(i) Write down the value of \(a\) and the value of \(b\).
(ii) \(z_2=-1+2i\). Plot \(z_2\) on the Argand Diagram.
(iii) \(z_3=\dfrac{z_1}{z_2}\). Write \(z_3\) in the form \(x+yi\), where \(x,y\in\mathbb{R}\).
(i) \(a=3\) and \(b=-1\)
(ii)
(iii) \(-1-i\)
(i) \(a=3\) and \(b=-1\)
(ii)
(iii)
\begin{align}z_3&=\frac{z_1}{z_2}\\&=\frac{3-i}{-1+2i}\\&=\frac{3-i}{-1+2i}\times\frac{-1-2i}{-1-2i}\\&=\frac{-3-6i+i-2}{1+2i-2i+4}\\&=\frac{-5-5i}{5}\\&=-1-i\end{align}
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(b) Solve for \(z\)
\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}
\(z=11-15i\)
\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2z-24+36i=-4+2i+4i+2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2z&=-4+2i+4i+2+24-36i\\&=22-30i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}z=11-15i\end{align}
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2016 Paper 1 Question 2
\(z_1=1+3i\) and \(z_2=2-i\), where \(i^2=-1\), are two complex numbers.
(a) Let \(z_3=z_1+2z_2\). Find \(z_3\) in the form ܽ\(a+bi\) where \(a,b\in\mathbb{Z}\).
\(z_3=5+i\)
\begin{align}z_3&=z_1+2z_2\\&=(1+3i)+2(2-i)\\&=1+3i+4-2i\\&=5+i\end{align}
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(b) Plot \(z_1\), \(z_2\) and \(z_3\) on the given Argand diagram and label each point clearly.
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(c) Investigate if \(|z_2-z_3|=|z_1+z_2|\).
\(|z_2-z_3|=|z_1+z_2|\)
\begin{align}|z_2-z_3|&=|(2-i)-(5+i)|\\&=|-3-2i|\\&=\sqrt{(-3)^2+(-2)^2}\\&=\sqrt{13}\end{align}
and
\begin{align}|z_1+z_2|&=|(1+3i)+(2-i)|\\&=|3+2i|\\&=\sqrt{3^2+2^2}\\&=\sqrt{13}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|z_2-z_3|=|z_1+z_2|\end{align}
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(d) Find the complex number \(w\), such that \(w=\frac{z_1}{z_2}\).
Give your answer in the form \(a+bi\) where \(a,b\in\mathbb{R}\).
\(w=-\dfrac{1}{5}+\dfrac{7}{5}i\)
\begin{align}w&=\frac{z_1}{z_2}\\&=\frac{1+3i}{2-i}\\&=\frac{1+3i}{2-i}\times\frac{2+i}{2+i}\\&=\frac{2+i+6i-3}{4+2i-2i+1}\\&=\frac{7i-1}{5}\\&=-\frac{1}{5}+\frac{7}{5}i\end{align}
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2015 Paper 1 Question 6
(a) The complex number \(z_1=a+bi\), where \(i^2=-1\), is shown on the Argand diagram below.
(i) Write down the value of a and the
value of b.
(ii) The image of \(z_1\) under reflection in the real axis is \(z_2=c+di\). Write down the value of \(c\) and the value of \(d\).
(i) \(a=3\) and \(b=1\)
(ii) \(c=3\) and \(d=-1\)
(i)
\begin{align}a=3&&b=1\end{align}
(ii)
\begin{align}c=3&&d=-1\end{align}
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(b)
(i) The angle \(\theta\) is formed by joining \(z_1\) to \(0+0i\) to \(z_2\).
Find \(\cos \theta\), correct to one decimal place.
(ii) Show that \(|z_1|\times|z_2|\times \cos \theta=ac+bd\), where \(a\), \(b\), \(c\) and \(d\) are the values from part (a) above.
(i) \(0.8\)
(ii) The answer is already in the question!
(i)
\begin{align}|z_1|&=\sqrt{3^2+1^2}\\&=\sqrt{10}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2^2=(\sqrt{10})^2+(\sqrt{10})^2-2(\sqrt{10})(\sqrt{10})\cos\theta\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos\theta&=\frac{10+10-4}{2(10)}\\&=0.8\end{align}
(ii)
\begin{align}|z_1|\times|z_2|\times\cos\theta&=\sqrt{10}\times\sqrt{10}\times0.8\\&=8\end{align}
and
\begin{align}ac+bd&=(3)(3)+(1)(-1)\\&=8\end{align}
as required.
