IB Math HL – Complex Numbers

1. Complex Numbers, Equality of complex numbers – IB Mathematics HL

For what values of x and y, the following complex numbers are equal

$z=(x+4)^2+3i$ and $w=16+yi$

Solution

Two complex numbers are equal if their corresponding real parts and imaginary parts are equal.

Therefore, $z=(x+4)^2+3i=16+yi=w \Rightarrow \ (x+4)^2=16$ and $y=3$

$\Rightarrow \ (x+4) = \pm 4$ and $y=3 \Rightarrow \ x+4=4 \ or\ x+4=-4 \ and\ y=3$

$\Rightarrow x=0 \ or\ x=-8 \ and\ y=3$

2. Complex Numbers, Operations with complex numbers – IB Mathematics HL

How can we find $z+2w \ ,\ z\cdot w \ ,\ z^2$ given that

$z=3+2i$ and $w=5-2i$

Solution

The sum (or the difference) of two complex numbers is the complex number which its real part is made up of the sum (or the difference) of their real parts and its imaginary parts are made up of the sum (or the difference) of their imaginary parts.

$z \pm w=(a+bi) \pm (c+id)=(a \pm c)+(b \pm d)i$

Therefore, $z+w=(3+2i)+(5-2i)=(3+5)+(2-2)i=8+0i=8$

The multiplication is performed as usual and using the fact that $i^2=-1$ we have the following:

$z \cdot w =(3+2i) \cdot (5-2i)=15-6i+10i-4i^2=19+4i$ and

$z^2=(3+2i)^2=3^2+12i+(2i)^2=9+12i-4=5+12i$

3. Complex Numbers, Division of Complex Numbers – IB Mathematics HL

How can we find $\frac{z}{w}$ given that

$z=3+4i$ and $w=5-2i$

Solution

When dividing two complex numbers, we multiply the numerator and denominator by the conjugate of the denominator such as the product in the denominator to become a real number.

$\frac{z}{w}= \frac{a+ib}{c+di}= \frac{a+ib}{c+di} \cdot \frac{c-di}{c-di}$

$}= \frac{(a+ib)(c-di)}{(c^2-(di)^2)}$

$}= \frac{(ac-adi+bci+bd)}{(c^2+d^2)}$

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

Thus, about this question we have that

$}= \frac{(3+4i)(5+2i)}{(5^2-(2i)^2)}$

$}= \frac{(7+26i)}{29}$

$}= \frac{7}{29} + \frac{26}{29} \cdot i$

4. Complex Numbers, Cartesian to polar form – IB Mathematics HL

How can we express the complex number $z=1- \sqrt3 i$ in polar form?

Solution

The argument $\theta$ of a complex number $z=a+ib$ is given by the formula:

$\theta = arctan(\frac{b}{a}$ and the modulus $|z| =\sqrt{a^2+b^2}$ and the polar form will be $z= |z|(cos \theta +isin \theta)= |z|cis( \theta)$

in this question

$\theta = arctan(\frac{-\sqrt3}{1}= -\frac{\pi}{3}$

and the modulus is given by the following formula:

$|z| =\sqrt{1^2+(-\sqrt3)^2}$

$|z| =\sqrt{1+3}=\sqrt{4}=2$

$z= |z|(cos ( -\frac{\pi}{3}) +isin( -\frac{\pi}{3}))= 2cis(- \frac{\pi}{3})$

5. Complex Numbers, Cartesian to polar form – IB Mathematics HL

How can we express the complex number $z=5i$ in polar form?

Solution

In this case the argument $\theta$ of the complex number $z=5i$ is $\theta = \frac{\pi}{2}$ and the modulus is

$|z| =\sqrt{5^2}=5$

So, the complex number $5i$ can be written in polar form as follows:

$z= 5(cos ( \frac{\pi}{2}) +isin( \frac{\pi}{2}))= 5cis( \frac{\pi}{2})$

6. Complex Numbers, the Cartesian form of Complex Numbers – IB Maths HL

How can we express the following complex number $z= 5(cos ( \frac{\pi}{6}) +isin( \frac{\pi}{6}))$ in polar form?

Solution

Complex Numbers, the Cartesian form of Complex Numbers – IB Maths HL

$z= 5(cos( \frac{\pi}{6}) +isin( \frac{\pi}{6}))$

$z= 5cos( \frac{\pi}{6}) +5isin( \frac{\pi}{6})$

$z= 5 \cdot \frac{\sqrt{3}}{2}) +5 \cdot \frac{1}{2} i$

$z= 5 \sqrt{3} + \frac{5}{2}i$

7. De Moivre’s Theorem – IB Mathematics HL

How can we find the following complex number

$z=(1- \sqrt3 i)^{24}$ using De Moivre’s Theorem?

Solution

De Moivre’s Theorem – IB Math HL

The De Moivre’s theorem is given by the following formula

$z^n=( |z|(cos \theta +isin \theta))^n$

$= |z|^n (cos \theta +isin \theta)^n$

$= |z|^n (cos (n \theta) +isin (n \theta))$

Before apply De Moivre’s theorem we must convert the Cartesian form to polar form

$z=(1- \sqrt3 i)=$

$z= 2(cos ( -\frac{\pi}{3}) +isin( -\frac{\pi}{3}))$

Therefore, $(1-\sqrt3 \cdot i)^ {24}=$

$=(2(cos(-\frac{\pi}{3})+isin(-\frac{\pi}{3})))^{24}$

$=2^{24} (cos(-24 \frac{\pi}{3}) +isin(-24 \frac{\pi}{3}))$

$=2^{24} (cos(-8 \pi) +isin(-8 \pi))$

$=2^{24} (1) +0 i =2^{24}$

8. Complex Numbers, Quadratic over the complex field – IB Mathematics HL

How can we solve the following quadratic equation over the complex field?

$z^2-6z+13=0$

SOL

The Discriminant (Δ) of the quadratic equation will be

$Δ=(-6)^2-4 (13)=36-52=-16=16i^2$

So, the two roots will be given by the following formula

$z_{1,2}= \frac{6 \pm \sqrt{16i^2}}{2}$

$z_{1,2}= \frac{6 \pm (4i)}{2}$

$z_{1,2}= 3 \pm 2i$