# HARD QUESTIONS ON COMPLEX NUMBERS

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

$2x^2-2ix-5=0$

Let's solve this using the quadratic formula.

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

$x=\frac{-b±\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2i)±\sqrt{(-2i)^2-4(2)(-5)}}{2(2)}$

$x=\frac{2i±\sqrt{4i^2+40}}{4}$

If $i^2=-1$, then

$x=\frac{2i±\sqrt{4(-1)+40}}{4}$

$x=\frac{2i±\sqrt{-4+40}}{4}$

$x=\frac{2i±\sqrt{36}}{4}$

$x=\frac{2i ± 6}{4}$

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:

$(x+iy)(2+i)=3-i$

$x(2+i)+iy(2+i)=3-i$

$2x+xi+2yi+yi^2=3-i$

If $i^2=-1$, then

$2x+xi+2yi+y(-1)=3-i$

$2x+xi+2yi-y=3-i$

$2x-y+(x+2y)i=3-(1)i$

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

$2x-y=3$

$x+2y=-1$

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

$x=-1-2y$

Inserting the value of $x$ in Eqn 1

$2(-1-2y)-y=3$

$-2-4y-y=3$

$-2-5y=3$

$-5y=3+2$

$-5y=5$

$\frac{-5y}{-5}=\frac{5}{-5}$

$y=-1$

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

$2x-(-1)=3$

$2x+1=3$

$2x=3-1$

$2x=2$

$x=1$

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

Question 3

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

Multiplying the complex numbers

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

$3x+4ix+3yi+4yi^2=3-4i$

$3x+4xi+3yi+4y(-1)=3-4i$

$3x+4xi+3yi-4y=3-4i$

$3x-4y+4xi+3yi=3-4i$

Equating the real and imaginary parts

$3x-4y=3$

$4x+3y=-4$

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

$4(3x-4y=3)$

$3(4x+3y=-4)$

$12x-16y=12$

$12x+9y=-12$

Subtracting both eqn have:

$-25y=24$

$\frac{-25y}{-25}=\frac{24}{-25}$

$y=\frac{-24}{25}$

Inserting the value of y in Eqn 1

$3x-4(\frac{-24}{25})=3$

$3x+\frac{96}{25}=3$

Taking the L.C.D

$\frac{75x+96}{25}=3$

$75x+96=75$

$75x=75-96$

$75x=-21$

$\frac{75x}{75}=\frac{-21}{75}$

$x=\frac{-7}{25}$

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:

$\frac{xy}{x}=y$

By substitution

$\frac{-14+2i}{-3+4i}=y$

To divide complex numbers, you simply rationalize the denominator

$\frac{-14+2i}{-3+4i}\times\frac{-3-4i}{-3-4i}=y$

$\frac{(-14+2i)(-3-4i)}{(-3+4i)(-3-4i)}$

$\frac{-14(-3-4i)+2i(-3-4i)}{-3(-3-4i)+4i(-3-4i)}$

$\frac{42+56i-6i-8i^2}{9-12i+12i-16i^2}$

If $i^2=-1$, then

$\frac{42+50i+8}{9+16}$

$\frac{50+50i}{25}$

$\frac{50}{25}+\frac{50i}{25}$

$2+2i$

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Question 5

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

$\frac{x}{y}=6-8i$

Substituting the value of $x$

$\frac{22+4i}{y}=6-8i$

Now, let's cross multiply

$22+4i=y(6-8i)$

$\frac{22+4i}{6-8i}=\frac{y(6-8i)}{6-8i}$

$\frac{22+4i}{6-8i}=y$

As usual, we rationalize the denominator

$y=\frac{22+4i}{6-8i}\times\frac{6+8i}{6+8i}$

$y=\frac{22(6+8i)+4i(6+8i)}{6(6+8i)-8i(6+8i)}$

$y=\frac{132+176i+24i+32i^2}{36+48i-48i-64i^2}$

$y=\frac{132+200i+32(-1)}{36-64(-1)}$

$y=\frac{132-32+200i}{36+64}$

$y=\frac{100+200i}{100}$

Splitting the fraction

$y=\frac{100}{100}+\frac{200i}{100}$

$y=1+2i$

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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