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Past Papers

## 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

## 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)

Solution

(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).

Solution
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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 $$|z_2-z_3|=\sqrt{106}$$

Solution

\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.

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.

Solution
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(b) $$z_3=\dfrac{z_1}{z_2}$$. Find $$z_3$$ in the form $$a+bi$$, where $$a,b\in\mathbb{Z}$$.

$$z_3=i$$

Solution

\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 $$q\in\mathbb{N}$$

$$7\sqrt{2}$$

Solution

\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) $$5$$

Solution

(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 $$a,b\in\mathbb{R}$$.

$$3+5i$$

Solution

\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.

Solution

(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 $$z^2-4z+4=0$$.

Solution

\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.

Solution

\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 $$|z_3|=|z_1|+|z_2$$.

$$|z_3|\neq|z_1|+|z_2|$$

Solution

\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}

Solution

\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) $$-1-i$$

Solution

(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}

$$z=11-15i$$

Solution

\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 $$a,b\in\mathbb{Z}$$.

$$z_3=5+i$$

Solution

\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.

Solution
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(c) Investigate if $$|z_2-z_3|=|z_1+z_2|$$.

$$|z_2-z_3|=|z_1+z_2|$$

Solution

\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 $$a,b\in\mathbb{R}$$.

$$w=-\dfrac{1}{5}+\dfrac{7}{5}i$$

Solution

\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 $$d=-1$$

Solution

(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) $$0.8$$

Solution

(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.

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