Solving Equations in Complex Numbers

Solving complex number equations. A definite approach to unsolvable

equations involving complex numbers It is assumed in the slides that you are familiar with De Moiver’s theorem and basics of factorization. We will see how we should use De Moiver’s

while finding z.

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Equations in Complex Numbers

Remember two most important thing in complex numbers while solving equations.

• If two complex numbers are equal, that implies their modulus and arguments are equal

• If two complex numbers are equal, that implies real and imaginary parts are equal.


Equations: Solving ProblemsThe whole problem of solving equations of complex numbers so depends on the

mentioned two facts. So, we need to focus to come to a point where we could equate two complex numbers. This is how we proceed while solving complex number equations:

• Factorize the equation as better as you can to make it of the form (z-z1)(z-z2)…. = 0. If this is not possible, try factorizing in form (zn1 – z1)(zn2 – z2) (zn3 – z3)….. = 0.

• If information about real part or imaginary part of z is asked, put z = x+iy and as you do this, equate real part and imaginary part of z . Else make a habit of putting z = rcis(φ) = rei φ and as you do this, finally equate modulus and argument of complex numbers. We will understand them better while doing problems. Not that if z is complex, then az + ibz = x+iy will not mean that az = x, and bz = y as we are not equating real part and imag. part since az can be itself complex too and may have i term. Make sure, real and imag parts which you are equating are real.

• Once you have come to this point we will see in next slide how to proceed the problem.


Equations: Solving Problems

An important thing to remember• Suppose you get einθ = eiα .This does not mean

that θ = α/n only but instead we have to apply De Moiver’s theorem here to conclude that θ = α/n, (2π+α)/n, (4π+α)/n, ……………………, (2(n-1)π+α)/n and has thus n solutions.

• z2 = i has two solutions. Equating modulus, we get |z| = 1 and since z = |z| eiθ , we will get ei2θ

= ei π/2 giving θ = (π/2) /2, (2π+ π/2)/2


Equations: Examples• Solve equation z3 + z = 0

• Note that as discussed, first step should be factorize the equation as far as we can. Here we can see that equation is nothing but z(z2+1) = 0 which gives z = 0 or z2+1. Straight away second equation would mean z = + or – iota.

But how do we solve such equation in general. We do as below. Since z2+1 = 0, => z2 = -1. Two complex numbers are equal implies their modulus is equal. So, for this case |z| = 1 and if z = reiθ then r = 1 leaving us with ei2θ =- 1 = e-iπ whose solutions as we know are

θ = (-π) /2, (2π+ (-π))/2


Equations : Examples• Solve for z z3 + iz2 – iz + 1 = 0. Again as mentioned, we should try to factorize the term. The crux of this problem is to

factorize the equation. • First of all write the equation in decreasing order of powers of z. Equation given is

already sorted. • Then take something common from first two terms in such a way that coefficient

of highest term of z in bracket becomes 1. e.g., in this case we will simple take z2 common so that first two terms become z2(z+i).

• Next remaining two terms are –iz + 1. Note that if it has to factorize, you should take something common such that coefficient of z in bracket should become 1. This hints that that in next two terms we have to take –i common. Now, we have to see last term and figure out what happen if we take –i from 1. Clearly as -i2 = 1, so taking –i common from +1 will leave us with i only and next two terms will become –iz(z+i) and that how it should be if problem was solvable. So the equation now becomes z2(z+i) – iz(z+i) = 0 or (z2 – i )(z+i) = 0.


Equations : Examples

• Equation we simplified is (z2 – i )(z+i) = 0.• The equation implies either z = -i or z2 = i. • Lets solve z2 = i. If we put z = reiθ then, z2 =

r2ei2θ. So r2ei 2θ = i. Equating modulus, we obtain r = 1 and this leaves us with ei 2θ = ei π/2

giving θ = (π/2) /2, (2π+ π/2)/2.


Equations : ExamplesNow, go to discuss problems sections and browse to

Mathematics > Algebra. Do the problems titled Solving Equations in Complex Number 1 , 2 , 3 etc.

Please ask doubts in the same forum. Coming weekend, we will learn how to do problems involving inequalities in complex numbers.

Also note that in majority of problems of such sort asked in IIT-JEE, only modulus is asked which as you saw is the easiest part as it involves just factorization and equating modulus on both sides.


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Solving Equations in Complex Numbers