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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)**

## 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.