Solving Equations in Complex Numbers

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Information about Solving Equations in Complex Numbers
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Published on February 24, 2009

Author: youmarks

Source: slideshare.net

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Various methods of solving problems in complex numbers exploiting De Moiver's Theorem

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. Brought to you by Youmarks Preparations Prepared by Parag Arora copyrights © youmarks.com

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. copyrights © youmarks.com

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 Problems The 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 (z n1 – z1)(z n2 – z2) (z n3 – 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 ) = re i 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. copyrights © youmarks.com

The 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 (z n1 – z1)(z n2 – z2) (z n3 – 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 ) = re i 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 e in θ = e i α .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, (2 π + α )/n, (4 π + α )/n, ……………………, (2(n-1) π + α )/n and has thus n solutions. z 2 = i has two solutions. Equating modulus, we get |z| = 1 and since z = |z| e i θ , we will get e i2 θ = e i π /2 giving θ = ( π /2) /2, (2 π + π /2)/2 copyrights © youmarks.com

An important thing to remember

Suppose you get e in θ = e i α .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, (2 π + α )/n, (4 π + α )/n, ……………………, (2(n-1) π + α )/n and has thus n solutions.

z 2 = i has two solutions. Equating modulus, we get |z| = 1 and since z = |z| e i θ , we will get e i2 θ = e i π /2 giving θ = ( π /2) /2, (2 π + π /2)/2

Equations: Examples Solve equation z 3 + 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(z 2 +1) = 0 which gives z = 0 or z 2 +1. Straight away second equation would mean z = + or – iota. But how do we solve such equation in general. We do as below. Since z 2 +1 = 0, => z 2 = -1. Two complex numbers are equal implies their modulus is equal. So, for this case |z| = 1 and if z = re i θ then r = 1 leaving us with e i2 θ =- 1 = e -i π whose solutions as we know are θ = (- π ) /2, (2 π + (- π ))/2 copyrights © youmarks.com

Solve equation z 3 + 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(z 2 +1) = 0 which gives z = 0 or z 2 +1. Straight away second equation would mean z = + or – iota.

But how do we solve such equation in general. We do as below.

Since z 2 +1 = 0, => z 2 = -1. Two complex numbers are equal implies their modulus is equal. So, for this case |z| = 1 and if z = re i θ then r = 1 leaving us with e i2 θ =- 1 = e -i π whose solutions as we know are

θ = (- π ) /2, (2 π + (- π ))/2

Equations : Examples Solve for z z 3 + iz 2 – 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 z 2 common so that first two terms become z 2 (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 -i 2 = 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 z 2 (z+i) – iz(z+i) = 0 or (z 2 – i )(z+i) = 0. copyrights © youmarks.com

Solve for z

z 3 + iz 2 – 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 z 2 common so that first two terms become z 2 (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 -i 2 = 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 z 2 (z+i) – iz(z+i) = 0 or (z 2 – i )(z+i) = 0.

Equations : Examples Equation we simplified is (z 2 – i )(z+i) = 0. The equation implies either z = -i or z 2 = i. Lets solve z 2 = i. If we put z = re i θ then, z 2 = r 2 e i2 θ . So r 2 e i 2 θ = i. Equating modulus, we obtain r = 1 and this leaves us with e i 2 θ = e i π/2 giving θ = ( π /2) /2, (2 π + π /2)/2. copyrights © youmarks.com

Equation we simplified is (z 2 – i )(z+i) = 0.

The equation implies either z = -i or z 2 = i.

Lets solve z 2 = i. If we put z = re i θ then, z 2 = r 2 e i2 θ . So r 2 e i 2 θ = i. Equating modulus, we obtain r = 1 and this leaves us with e i 2 θ = e i π/2 giving θ = ( π /2) /2, (2 π + π /2)/2.

Equations : Examples Now, 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. copyrights © youmarks.com

Now, 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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