Cubic EquationsThe Imaginary Number and Transcendental Functions
The imaginary number, i, defined as \(\sqrt{-1}\), is a critical number in mathematics, although it has no real value. Although the number itself has no value, applying some functions or calculations can result in real numbers. For example, the powers of i alternate between real and imaginary numbers:
(1) \(i^1 = i\) and \(i^{4n+1} = i\)
(2) \(i^2 = -1\) and \(i^{4n+2} = -1\)
(3) \(i^3 = -i\) and \(i^{4n+3} = -i\)
(4) \(i^4 = 1\) and \(i^{4n} = 1\) for all integers n > 0
This page contains various calculations with i. The goal is to represent various calculations as a complex number a + bi, where a is a real number and bi is the imaginary number.
First, recall that if a polynomial with integer coefficients has imaginary roots, they always occur as conjugate pairs. Therefore, if a + bi is a root, then a – bi is a root. This is the only way the imaginary part is eliminated to ensure the coefficients are real. The product of the conjugate pairs is real: \((a + bi)(a - bi) = \) \(a^2 + abi - abi - b^2\cdot i^2 = \) \(a^2 + b^2\).
First, we’ll list some identities and then offer some proofs of some of them.
Exponential Functions
Euler first noted the link between exponential functions and trigonometric functions when he discovered the identity: \(e^{xi} = \cos x + i\cdot\sin x\). We can generalize this for e raised to any complex number:
(1) \(e^{a+bi} = e^a\cdot e^{bi} = e^a(\cos b + i\cdot\sin b)\)
Equation (1) gives the desired result as a + bi. We can also take the logarithm of a complex number:
(2) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \( i\cdot \arctan\dfrac{b}{a}\)
If a = 0, then (2) becomes \(\ln(bi) = \ln b + \dfrac{\pi}{2}i\). Here are some other values:
(3) \(i^i = e^{-\pi/2}\)
Note that the imaginary unit raised to the imaginary unit becomes a real number.
(4) \(x^i = e^{(\ln x)i} = \) \( \cos(\ln x) + i\cdot\sin(\ln x)\)
(5) \((a+bi)^i = \) \( e^{-\arctan(b/a)}\times\) \(\left[\cos(\ln\sqrt{a^2+b^2}) + i\cdot\sin(\ln\sqrt{a^2+b^2}) \right]\)
In equation (5), if we let b = 0, we get equation (4). If we let a = 0, and allow arctan(∞) = π/2, then we get \((bi)^i = e^{-\pi/2}\left[\cos(\ln b) + i\cdot\sin(\ln b) \right]\). If we let b = 1, then we get equation (3).
(5a) \(x^{a+bi} = x^a[\cos(b\ln x) + i\sin(b\ln x)]\)
Trigonometric Functions
(6) \(\sin(a + bi) = \) \(\sin a\cdot\cosh b + \) \( i\cdot\cos a\cdot\sinh b\)
(7) \(\cos(a + bi) = \) \(\cos a\cdot\cosh b - \) \( i\cdot\sin a\cdot\sinh b\)
(8) \(\tan(a+bi) = \) \(\dfrac{\tan a + \tan(bi)}{1-\tan a\cdot\tan(bi)} = \) \(\dfrac{\tan a\cdot\text{sech}^2 b+i\cdot\tanh b\cdot\sec^2 a}{1+(\tan a\cdot\tanh b)^2}\)
In equation (8), if we split the denominator between the terms of the numerator, we do get the a + bi form.
(9) \(\arcsin(a+bi) = \) \( -i\ln\left[\sqrt{1-(a+bi)^2}+i(a+bi) \right]\)
Well, equation (9) doesn’t help much since it’s not in the a + bi form. I’ll have to look at this one again. However, this is the equivalence given by Wolfram Alpha.
(10) \(\arcsin a = \dfrac{\pi}{2} - i\cosh^{-1} a\) (for all real a)
(11) \(\arccos(a+bi) = \) \(\dfrac{\pi}{2} + i\ln\left[\sqrt{1-(a+bi)^2}+i(a+bi) \right]\)
(12) \(\arccos a = i\cosh^{-1} a\) (for all real a)
(12a) \( \arctan(±x+yi)=±\frac{\pi}{4} + \frac{1}{2}\arctan\left(\frac{x^{2}+y^{2}-1}{2|x|}\right) + \frac{i}{4}\ln\left(\frac{x^2+(y+1)^2}{x^2+(y-1)^2}\right) \)
Hyperbolic Functions
(13) \(\sinh(a+bi) = \) \(\sinh a\cdot\cos b + \) \(i\cdot\cosh a\cdot\sin b\)
(14) \(\cosh(a+bi) = \) \(\cosh a\cdot\cos b - \) \( i\cdot\sinh a\cdot\sin b\)
(15) \(\tanh(a+bi) = \) \( \dfrac{\tanh a + \tanh(bi)}{1+\tanh a\cdot\tanh(bi)} = \) \( \dfrac{\tanh a\cdot\text{sec}^2 b+i\cdot\tan b\cdot\text{sech}^2 a}{1+(\tanh a\cdot\tan b)^2}\)
(16) \(\sinh^{-1}(a+bi) = \) \(\ln\left[\sqrt{(a+bi)^2+1}+a+bi \right]\)
(17) \(\cosh^{-1}(a+bi) = \) \(\ln\left[\sqrt{(a+bi)^2-1}+a+bi \right]\)
Algebraic Operations
(18) \( \sqrt{a + bi} = \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}} + i\sqrt{\frac{{\sqrt{a^{2}+b^{2}}-a}}{2}} \)
(19) \((a+bi)^n = \) \((\sqrt{a^2+b^2})^n\cdot \) \((\cos(n\theta) + i\cdot\sin(n\theta))\), where θ = arctan(b/a)
(20) \((a+bi)^{1/n} = \) \( (\sqrt{a^2+b^2})^{1/n}\cdot \) \( \left [\cos\left(\dfrac{\theta+2k\pi}{n}\right) + i\cdot\sin\left(\dfrac{\theta+2k\pi}{n}\right) \right ]\), where θ = arctan(b/a) and k = 0, 1, 2, ..., (n – 1)
Equation (20) gives all the nth roots of a complex number since any given number has n number of roots. Let’s look at some examples.
Example 1
Equations (19) and (20) are unique in that they allow us to find the nth power or nth roots of a complex number. We can solve some problems with these.
Problem: Evaluate (1 + i)1000
First, θ = arctan(1/1) = π/4. The quantity \(\sqrt{a^2+b^2}\) is the radius, r, of the complex number plotted on a polar graph (or the distance from the origin). For our example, \(r=\sqrt{1^2+1^2} = \sqrt{2}\). Putting this together:
(i) \((1+i)^{1000} = \) \((\sqrt{2})^{1000}\times\) \(\left[\cos(1000\pi/4) + i\cdot\sin(1000\pi/4)\right] \)
(ii) \((1+i)^{1000} = \) \(2^{500}\times \) \(\left[\cos(250\pi) + i\cdot\sin(250\pi)\right] \)
The cosine of odd multiples of π is –1 and even multiples is 1. The sine of both odd or even multiples of π is 0. Therefore,
(iii) \((1+i)^{1000} = 2^{500}\cdot(1 + 0) = 2^{500} \)
An alternative quick way of solving this is to square the number first and see if it simplified to something simple. Then we can raise it to the 500th power.
(i) \( (1+i)^{2} = 1 + 2i + i^{2} = 2i \)
(ii) \( (1+i)^{1000} = (2i)^{500} = 2^{500} \cdot i^{500} = 2^{500} \)
Since \( i^{500} = 1 \), the above simply simplified to 2500.
Example 2
Problem: Evaluate (2+2i)6
Since a and b are the same, we know that θ must be π/4. The radius is \(\sqrt{2^2+2^2} = 2\sqrt{2}\). Putting this together:
(i) \((2+2i)^{6} = \) \(\left(2\sqrt{2}\right)^6\times \) \(\left[\cos(6\pi/4) + i\cdot\sin(6\pi/4)\right] \)
(ii) \((2+2i)^{6} = \) \(512\cdot\left[\cos(3\pi/2) + i\cdot\sin(3\pi/2)\right] \)
Cosine of 3π/2 = 0 and sine of 3π/2 = –1. Therefore:
(iii) \((2+2i)^{6} = 512(0 - i) = -512i\)
If the complex number is of the form a + ai and it is raised to an even power, we can easily do this calculation. We first raise it to the second power, which eliminates the real part.
(i) \((a + ai)^{2k} =\) \([(a+ai)^2]^k\) (for k is any integer > 0)
(ii) \([a^2 + 2a^{2}i + (ai)^2]^k =\) \((2a^{2}i)^k\)
We can write (ii) as \((\sqrt{2}a)^{n}i^{n/2}\), where n is any even number. In example 2, our a = 2 and n = 6. So, \((2\sqrt{2})^{6}i^{3} = -512i\).
Example 3
Problem: Evaluate \(\left(1-\sqrt{3}i\right)^{2013}\)
The angle measure is \(\theta = \arctan\dfrac{-\sqrt{3}}{1} = -\dfrac{\pi}{3}\). In polar form, our complex number lies in the 4th quadrant. The distance from the origin is \(r = \sqrt{3 + 1} = 2\). Putting it together.
(i) \(\left(1-\sqrt{3}i\right)^{2013} = \) \(2^{2013}\times\) \(\left[\cos(-2013\pi/3) + i\cdot\sin(-2013\pi/3)\right]\)
(ii) \(\left(1-\sqrt{3}i\right)^{2013} = \) \(2^{2013}\times\) \(\left[\cos(-671\pi) + i\cdot\sin(-671\pi)\right]\)
Cosine of –671π = –1 because 671 is an odd multiple of π and sine of –671π = 0 because 671 is an odd multiple of π. Therefore:
(iii) \(\left(1-\sqrt{3}i\right)^{2013} = 2^{2013}(-1 + 0) = -2^{2013}\)
Roots of Complex Numbers
The Square Roots of i
The roots of complex numbers can be reduced to complex numbers of the form \( a + bi \). Complex numbers have n number of nth roots. So \( \sqrt[4]{i} \) would have 4 roots.
The roots can be found by using the relationship: \(a+bi = \sqrt{a^{2}+b^{2}}\cdot (\cos \theta + i\cdot\sin \theta)\) where θ is the angle made by the coordinate.
But before using this, let’s use a trick to find the square roots of -i and i. We know that \( \sqrt{i} \) can be represented as \(a+bi\). Let’s square both sides and solve for a and b to see where it leads us.
(i) \( (a+bi)^{2} =\left(\sqrt{i}\right)^{2} \)
(ii) \( a^{2} + 2abi + b^{2}i^{2} = i \)
(iii) \( a^{2} - b^{2} + 2abi = i \)
When we separate the real part from the imaginary part in (iii), we can see the \( 2abi = i \), and therefore, \(2ab = 1\). This means a and b must both be positive or both be negative.
From (iii), we can also see that \( a^{2} - b^{2} = 0 \) or \( a^{2} = b^{2} \) or \( a = b \). Substituting this into \(2ab = 1\), we get \( 2a^{2} = 1 \) or \(a = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2} \). And since both a and b are equal, it follows that the square roots of i are \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \) and \( -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \).
If we square \( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \), we get:
(iv) \( \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \right)^{2} = \frac{2}{4} -2\cdot\frac{2}{4} \cdot i + \frac{2}{4}\cdot i^{2}\)
(v) \( \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \right)^{2}= -i \)
So the conjugate \( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \) is the square root of \( -i \). And the other root of \( -i \) is \( -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \).
The Cube Roots of i
Let’s try the same process to find the cube roots of i.
(i) \( (a+bi)^{3} =\left(\sqrt[3]{i}\right)^{3} \)
(ii) \( a^{3}+ 3a^{2}bi + 3a(bi)^{2} + (bi)^{3} = i \)
(iii) \( a^{3} - 3ab^{2} + (3a^{2}b - b^{3})i = i \)
Again, separating the real part and the imaginary part in (iii), we have the two relationships \( a^{3} - 3ab^{2} = 0 \) and \( 3a^{2}b - b^{3} = 1\).
From the first relationship \( a^{3} - 3ab^{2} = 0 \), we can easily find that a = 0 is one solution for a. When a = 0, we have the following in the second relationship:
(iv) \( 0 - b^{3} = 1\)
(v) \( b^{3} = -1\)
(vi) \( b = -1\)
Since a = 0 and b = -1, we have our first solution: \( 0 - 1\cdot i = -i\). So, -i is one cube roots of i, and (-i)3 does equal i.
Now, let’s work with the first relationship again and solve for b.
(vii) \( a^{3} - 3ab^{2} = 0 \)
(viii) \( a^{2} - 3b^{2} = 0 \)
(ix) \( a^{2} = 3b^{2} \)
Now we have a in terms of b. We can sub this into the second relationship.
(x) \( 3a^{2}b - b^{3} = 1 \)
(xi) \( 3\cdot 3b^{2}\cdot b - b^{3} = 1 \)
(xii) \( 8b^{3} = 1 \)
(xi) \( b = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \)
There is only 1 solution for b. Back substitution gives the solution for a: \( a = \pm\frac{\sqrt{3}}{2} \).
Therefore, the cube roots of i are: \( -i \), \( \frac{\sqrt{3}}{2} + \frac{1}{2}i \), and \( -\frac{\sqrt{3}}{2} + \frac{1}{2}i \).
The Square Roots of \(a+bi\)
Since \( a + bi = \sqrt{a^{2}+b^{2}}\cdot e^{\theta i} = \sqrt{a^{2}+b^{2}}(\cos{\theta} + i\sin{\theta}) \), \( \sqrt{a + bi} = \sqrt[4]{a^{2}+b^{2}}(\cos{\frac{1}{2}\theta} + i\sin{\frac{1}{2}\theta}) \), where \( \theta = \tan^{-1}{\frac{b}{a}} \).
We can do the math to reduce this to get the \(a+bi\) form using the half-angle identity of sine and cosine.
(i) \( \sqrt{a + bi} = \sqrt[4]{a^{2}+b^{2}}\left[\cos{\left(\frac{1}{2}\tan^{-1}\frac{b}{a}\right)} + i\sin{\left(\frac{1}{2}\tan^{-1}\frac{b}{a}\right)}\right] \)
(ii) \( \sqrt{a + bi} = \sqrt[4]{a^{2}+b^{2}}\left[\frac{\sqrt{\sqrt{a^{2}+b^{2}}+a}}{\sqrt{2\sqrt{a^{2}+b^{2}}}} + i\frac{\sqrt{\sqrt{a^{2}+b^{2}}-a}}{\sqrt{2\sqrt{a^{2}+b^{2}}}}\right] \)
(iii) \( \sqrt{a + bi} = \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}} + i\sqrt{\frac{{\sqrt{a^{2}+b^{2}}-a}}{2}} \)
Equation in (iii) gives 1 root. The other root is the negative of that.
Example
Find the roots of \(3+4i\).
(i) \( \sqrt{3+4i} = \sqrt{\frac{\sqrt{3^{2}+4^{2}}+3}{2}} + i\sqrt{\frac{{\sqrt{3^{2}+4^{2}}-3}}{2}} \)
(ii) \( \sqrt{3+4i} = \sqrt{\frac{\sqrt{25}+3}{2}} + i\sqrt{\frac{{\sqrt{25}-3}}{2}} \)
(iii) \( \sqrt{3+4i} = \sqrt{\frac{8}{2}} + i\sqrt{\frac{2}{2}} \)
(iv) \( \sqrt{3+4i} = \sqrt{4} + i\sqrt{1} \)
(v) \( \sqrt{3+4i} = 2 + i \)
The roots of \(3+4i\) are \(2+i\) and \(-2-i\). That reduced nicely. I chose \(3+4i\) purposefully since 3-4-5 triangle is a Pythagorean triplet. It would seem all Pythagorean triplets would reduce to nice rational coefficients, but this is not so. For the roots of \(6+8i\), we get the following:
(i) \( \sqrt{6+8i} = \sqrt{\frac{\sqrt{6^{2}+8^{2}}+6}{2}} + i\sqrt{\frac{{\sqrt{6^{2}+8^{2}}-6}}{2}} \)
(ii) \( \sqrt{6+8i} = \sqrt{\frac{10+6}{2}} + i\sqrt{\frac{4}{2}} \)
(iii) \( \sqrt{6+8i} = 2\sqrt{2} + i\sqrt{2} \)
Link to Pythagorean Triplets
Although the square roots are not always rational numbers, the square of \(a + bi\) always produces a Pythagorean triplet if a and b are integers.
\( (a + bi)^{2} = a^{2} + 2abi + (bi)^{2} = a^{2} - b^{2} + 2abi \)
In the equation above \(a^{2} - b^{2} \) and \(2ab\) are the legs of a right triangle. The third side is:
\( (a^{2} - b^{2})^{2} + (2ab)^{2} = a^{4} - 2a^{2}b^{2} + b^{4} + 4a^{2}b^{2} = (a^{2} + b^{2})^{2} \)
Proofs
Natural Logarithm
Proof of equation (2). First, we convert a + bi to its trigonometric form: \(a+bi = \sqrt{a^2+b^2}(\cos\theta + i\cdot\sin\theta)\), where θ = arctan(b/a). Now, we apply the log to both equivalents:
(i) \(\ln(a+bi) = \) \(\ln\left[ \sqrt{a^2+b^2}(\cos\theta + i\cdot\sin\theta) \right]\)
(ii) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(\ln(\cos\theta + i\cdot\sin\theta)\)
From equation (1), we can rewrite \(\cos\theta + i\cdot\sin\theta\) as \(e^{\theta i}\). Therefore:
(iii) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(\ln(e^{\theta i}) =\) \(\dfrac{1}{2}\ln(a^2+b^2) + \theta i\)
We already defined θ = arctan(b/a). Substitution gives the final result.
(iv) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(i\cdot\arctan\dfrac{b}{a}\)
Acrtangent
The arctangent of \( x + yi \) can be expressed as \( a + bi \) by differentiating and integrating in terms of y. The method is outlined here:
(i) \( \frac{d}{dy} [\arctan(x+yi)] = \frac{i}{1+(x+yi)^{2}} \)
(ii) \( \int \arctan(x+yi) \, dy = \int \frac{i}{1+(x+yi)^{2}} \, dy \)
(iii) \( \int \arctan(x+yi) \, dy = \int \frac{i}{1+x^{2}-y^{2}+2xyi} \, dy\)
Multiply by the conjugate \( 1+x^{2} + y^{2}-2xyi \) to get the integrand into standard form:
(iv) \( \int \arctan(x+yi) \, dy = \int \frac{i(1 + x^{2} - y^{2}) - 2xyi^{2} }{(1+x^{2}-y^{2})^{2}+4x^{2}y^{2}} \, dy\)
(v) \( \int \arctan(x+yi) \, dy = \int \frac{2xy}{(1+x^{2}-y^{2})^{2}+4x^{2}y^{2}} \, dy + i\int \frac{1+x^{2}-y^{2}}{(1+x^{2}-y^{2})^{2}+4x^{2}y^{2}} \, dy\)
These 2 integrals are not easy to evaluate. I used a computer program, which gives the following:
(12a) \( \arctan(±x+yi)=±\frac{\pi}{4} + \frac{1}{2}\arctan\left(\frac{x^{2}+y^{2}-1}{2|x|}\right) + \frac{i}{4}\ln\left(\frac{x^2+(y+1)^2}{x^2+(y-1)^2}\right) \)
The constant C can be determined to be \( \frac{\pi}{4} \) for positive x and \( -\frac{\pi}{4} \) for negative x by letting x approach 0 from the positive axis for positive x and negative axis for negative x.