Complex Number Exercises: List of Solved Questions and Feedback

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You complex numbers make it possible to solve mathematical problems that do not have solutions in the set of real numbers.

In a complex number written as $\dpi{120} z = a+ bi$ , we say that $\dpi{120} to$ is the real part, $\dpi{120} b$ is the imaginary part and $\dpi{120} i =\sqrt{-1}$ it is the imaginary unit.

To perform operations with complex numbers, there are some expressions that make calculations easier. Consider $\dpi{120} z_1 = a+ bi$ and $\dpi{120} z_2 = c + di$ .

Addition expression between complex numbers:

$\dpi{120} z_1 + z_2= (a+c)+(b + d) i$

Expression of subtraction between complex numbers:

$\dpi{120} z_1 - z_2= (a-c)+(b - d) i$

Expression of multiplication between complex numbers:

$\dpi{120} z_1 \cdot z_2= (ac - db)+(ad +cb) i$

Expression of division between complex numbers:

$\dpi{120} \frac{z_1}{z_2}= \frac{(ac + bd)}{c^2 + d^2} + \frac{(bc - ad)}{c^2 + d^2 }i$

Below is a list of questions solved with exercises on complex numbers. Learn to use each of the concepts involving these numbers!

Index

List of exercises on complex numbers
Resolution of question 1
Resolution of question 2
Resolution of question 3
Resolution of question 4
Resolution of question 5
Resolution of question 6
Resolution of question 7
Resolution of question 8

List of exercises on complex numbers

Question 1. Considering the complex numbers $\dpi{120} z_1 = 2 + 3i$ , $\dpi{120} z_2 = 2 - 5i$ and $\dpi{120} z_3 = -1 + 4i$ determine the value of $\dpi{120} A$ , When $\dpi{120} A= z_2 +4z_3 -3z_1$ .

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Question 2. Find the values of $\dpi{120} x$ and $\dpi{120} y$ such that $\dpi{120} (2 +xi) + (y-5i) = 3-i$ .

Question 3. Considering the complex numbers $\dpi{120} z_1 = -2 - 5i$ and $\dpi{120} z_2 = 1 + 3i$ , determine the value of $\dpi{120} A\cdot B$ , When $\dpi{120} A=z_1\cdot \bar{z_1}$ and $\dpi{120} B=z_2\cdot \bar{z_2}$ .

Question 4. Calculate the value of $\dpi{120} p$ and $\dpi{120} q$ for what $\dpi{120} z_1: z_2 = q + 2i$ , When $\dpi{120} z_1 = 3 - pi$ and $\dpi{120} z_2 = 1 + 2i$ .

Question 5. Determine the value of $\dpi{120} to$ for what $\dpi{120} (a + 3i): (3 + 2i)$ be a pure imaginary number.

Question 6. Calculate the following imaginary unit powers $\dpi{120} i$ :

The) $\dpi{120} i^{16}$
B) $\dpi{120} i^{200}$
ç) $\dpi{120} i^{829}$
d) $\dpi{120} i^{11475}$

Question 7. Find the solution to the equation $\dpi{120} x^2 + 9 = 0$ in the set of complex numbers.

Question 8. Determine the solution of the equation $\dpi{120} x^2 + x + 1 = 0$ in the set of complex numbers.

Resolution of question 1

We have $\dpi{120} z_1 = 2 + 3i$ and $\dpi{120} z_2 = 2 - 5i$ and $\dpi{120} z_3 = -1 + 4i$ and we want to determine the value of $\dpi{120} A$ , When $\dpi{120} A= z_2 +4z_3 -3z_1$ .

First, let's calculate $\dpi{120} 4z_3$ and $\dpi{120} 3z_1$ , separately:

$\dpi{120} 4z_3 = 4.(-1 + 4i) = -4 + 16i$

$\dpi{120} 3z_1 = 3.(2 + 3i) = 6 + 9i$

Now let's calculate $\dpi{120} A$ :

$\dpi{120} A= z_2 +4z_3 -3z_1$

$\dpi{120} \Rightarrow A= (2 - 5i) +(-4+16i) -(6+9i)$

$\dpi{120} \Rightarrow A= (2-4-6) + (-5+16-9)i$

$\dpi{120} \Rightarrow A= -8 + 2i$

Resolution of question 2

We want to find x and y so that $\dpi{120} (2 +xi) + (y-5i) = 3-i$ .

By expression of the sum between two complex numbers, we have to:

$\dpi{120} (2 +xi) + (y-5i) = 3-i$

$\dpi{120} \Rightarrow (2 + y) + (x-5)i = 3-i$

So we must have $\dpi{120} (2 + y) = 3$ and $\dpi{120} (x-5)i=-i$ . Let's solve these two equations to find x and y.

$\dpi{120} (2 + y) = 3\Rightarrow y = 3-2\Rightarrow y =1$

$\dpi{120} (x-5)i=-i\Rightarrow x- 5 = -1 \Rightarrow x = -1 + 5 \Rightarrow x = 4$

Resolution of question 3

We have $\dpi{120} z_1 = -2 - 5i$ and $\dpi{120} z_2 = 1 + 3i$ and we want to determine the value of $\dpi{120} A\cdot B$ , When $\dpi{120} A=z_1\cdot \bar{z_1}$ and $\dpi{120} B=z_2\cdot \bar{z_2}$ .

First, we calculate $\dpi{120} A=z_1\cdot \bar{z_1}$ .

$\dpi{120} A=z_1\cdot \bar{z_1}$

$\dpi{120} \Rightarrow A = (-2 - 5i)\cdot (-2+5i)$

By the expression of the multiplication between two complex numbers, we have to:

$\dpi{120} A =[(-2)\cdot (-2) -(-5)\cdot 5 ]+[(-2)\cdot 5 + (-5)\cdot (-2)]$

$\dpi{120} \Rightarrow A =[4 +25]+[-10 +10]$

$\dpi{120} \Rightarrow A =29$

Now let's calculate $\dpi{120} B=z_2\cdot \bar{z_2}$ .

$\dpi{120} B=z_2\cdot \bar{z_2}$

$\dpi{120} \Rightarrow B = (1 + 3i)\cdot (1-3i)$

$\dpi{120} \Rightarrow B = [1\cdot 1 - 3\cdot (-3)] +[1\cdot (-3)+1\cdot 3]i$

$\dpi{120} \Rightarrow B = [1 + 9] +[-3+3]i$

$\dpi{120} \Rightarrow B = 10$

Therefore, $\dpi{120} A\cdot B = 29\cdot 10 = 290$ .

Resolution of question 4

We want to calculate the value of $\dpi{120} p$ and $\dpi{120} q$ for what $\dpi{120} z_1: z_2 = q + 2i$ , When $\dpi{120} z_1 = 3 - pi$ and $\dpi{120} z_2 = 1 + 2i$ .

It means finding $\dpi{120} p$ and $\dpi{120} q$ so that:

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$\dpi{120} \frac{3-pi}{1+2i} = q + 2i$

By the expression of the division between two complex numbers, we have to:

$\dpi{120} \frac{3-pi}{1+2i} = \frac{[3\cdot 1+(-p)\cdot 2]}{1^2+2^2} + \frac{[ (-p)\cdot 1-3\cdot 2]}{1^2+2^2}i = \frac{3 - 2p}{5} + \frac{(-p - 6)}{5}i$

Joining the two conditions, we must have:

$\dpi{120} \frac{3 - 2p}{5} + \frac{(-p - 6)}{5}i = q + 2i$

I.e:

$\dpi{120} \frac{3 - 2p}{5} = q \: \: \mathrm{e}\: \: \frac{(-p-6)}{5}i = 2i$

Let's solve each of these equations, starting with the second that only depends on p.

$\dpi{120} \frac{(-p-6)}{5}i = 2i$

$\dpi{120} \Rightarrow \frac{(-p-6)}{5} = 2$

$\dpi{120} \Rightarrow -p - 6 = 10$

$\dpi{120} \Rightarrow p = -16$

Now, we find q by the other equation:

$\dpi{120} \frac{3 - 2p}{5} = q$

$\dpi{120} \Rightarrow \frac{3 - 2\cdot (-16)}{5} = q$

$\dpi{120} \Rightarrow q = 7$

Resolution of question 5

We want to find the value of $\dpi{120} to$ for what $\dpi{120} (a + 3i): (3 + 2i)$ be a pure imaginary number.

A pure imaginary number is one whose real part is equal to zero.

Considering the expression of the division between two complex numbers, we have that:

$\dpi{120} \frac{a + 3i}{3 + 2i} = \frac{a\cdot 3 + 3\cdot 2}{3^3 + 2^2} + \frac{3\cdot 3 - a \cdot 2}{3^3 + 2^2}i = \frac{3a + 6}{13} + \frac{9-2a}{13}i$

For this number to be pure imaginary, we must have:

$\dpi{120} \frac{3a + 6}{13} = 0$

$\dpi{120} \Rightarrow 3a + 6 = 0$

$\dpi{120} \Rightarrow a = -2$

Resolution of question 6

By defining powers and complex numbers we have to:

$\dpi{120} i^0 = 1$
$\dpi{120} i^1 = i$
$\dpi{120} i ^2 = -1$
$\dpi{120} i^3 = -i$
$\dpi{120} i^4=1$
$\dpi{120} i^5 = i$
$\dpi{120} i^6 = -1$
$\dpi{120} i^7 = -i$

Observe a pattern that repeats itself every four successive powers: 1, i, -1 and -i.

Thus, to find the result at any power of i, just divide the exponent by 4. The remainder of the division will be 0, 1, 2 or 3 and this value will be the exponent we should use.

The) $\dpi{120} i^{16}$

16: 4 = 4 and the rest is 0.

Then, $\dpi{120} i^{16} = i^0 = 1$ .