We have already discussed addition, subtraction, multiplication, and division of complex numbers. Here are some question that will solidify your knowledge on complex numbers

Question 1

Solve the equation



Let's solve this using the quadratic formula.

$a=2$, $b=-2i$, $c=-5$.




If $i^2=-1$, then




$x=\frac{2i ± 6}{4}$

Splitting the plus and minus sign, we have 

$x=\frac{2i+6}{4}$ or $\frac{2i-6}{4}$

$x=\frac{\require{cancel}\bcancel{2}(i+3)}{\require{cancel}\bcancel{2}(2)}$ or $\frac{\require{cancel}\bcancel{2}(i-3)}{\require{cancel}\bcancel{2}(2)}$

$x=\frac{1-3i}{2}$ or $\frac{1+3i}{2}$

$x=\frac{1}{2}-\frac{3}{2}i$ or $\frac{1}{2}+\frac{3}{2}i$

Question 2

Find the value of $x$ and $y$ in the following equation:



Let's multiple this as we usually do for complex number



If $i^2=-1$, then




Now, let's equate real numbers to real numbers and imaginary number to imaginary numbers.



Let make $x$ the subject of the formula in Eqn 2


Inserting the value of $x$ in Eqn 1








Now, substituting $-1$ for y in Eqn 1






Therefore, $x$ and $y$ are $1$ and $-1$ respectively

Question 3

Solve $(x+iy)(3+4i)=3-4i$


Multiplying the complex numbers






Equating the real and imaginary parts



Solving simultaneously by multiplying eqn 1 by 4 and eqn 2 by 3





Subtracting both eqn have:




Inserting the value of y in Eqn 1



Taking the L.C.D







Question 4

If $x=-3+4i$ and $xy=-14+2i$, find the value of y


If we divide $xy$ by $x$, we get $y$. That is:


By substitution


To divide complex numbers, you simply rationalize the denominator





If $i^2=-1$, then





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Question 5

If $x=22+4i$ and $\frac{x}{y}=6-8i$, find $y$.



Substituting the value of $x$


Now, let's cross multiply




As usual, we rationalize the denominator







Splitting the fraction



Voila! we just conclude our series on complex numbers. If you have got questions relating to this, feel free to ask our telegram community

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