Method of completing squares in Complex Numbers The purpose of this slide is to show how do we complete squares in complex numbers. A very beautiful solution for finding square root of (a+ib) is also demonstrated. It is assumed that you know definition of iota and complex numbers Prepared By Parag Arora copyrights © youmarks.com

Problem Find z = One way to find this is by equation z to x + iy, on squaring which gives 7 + 24i = x 2 – y 2 + 2ixy x 2 – y 2 = 7 and xy = 12 which on solving gives the value of x and y. It must be emphasized here that if we visualize 7 = 4 2 – 3 2 and 12 = 4.3, we straight away get value of x and y. This is known as method of completing square. The method of completing squares is shown here for a particular case. We will however generalize this method using a very beautiful approach of completing squares. copyrights © youmarks.com

Find z =

One way to find this is by equation z to x + iy, on squaring which gives

7 + 24i = x 2 – y 2 + 2ixy

x 2 – y 2 = 7 and xy = 12 which on solving gives the value of x and y.

It must be emphasized here that if we visualize 7 = 4 2 – 3 2 and 12 = 4.3, we straight away get value of x and y. This is known as method of completing square.

The method of completing squares is shown here for a particular case. We will however generalize this method using a very beautiful approach of completing squares.

Finding Again as we did, one method is put x + iy = This would as before give us x 2 – y 2 =a and xy = b/2. We can easily solve the above two equations and find x and y in terms of a and b. But we will present a way of completing squares for finding the roots very fast. copyrights © youmarks.com

Again as we did, one method is put x + iy =

This would as before give us

x 2 – y 2 =a and xy = b/2.

We can easily solve the above two equations and find x and y in terms of a and b. But we will present a way of completing squares for finding the roots very fast.

Finding We know that ( c + id ) 2 = c 2 – d 2 + 2icd. To find We see that we need to figure out c and d in such a way that if cd = b/2 then c 2 – d 2 = a. Let us consider the case when b > 0. Now b = √b 2 Or copyrights © youmarks.com

We know that

( c + id ) 2 = c 2 – d 2 + 2icd. To find

We see that we need to figure out c and d in such a way that if cd = b/2 then c 2 – d 2 = a.

Let us consider the case when b > 0. Now

b = √b 2

Or

Finding Also note that a = copyrights © youmarks.com

Finding So we see that complex number a + ib is nothing but copyrights © youmarks.com

Finding So we see a + bi is is denoted by |z| if z = a + ib copyrights © youmarks.com

Finding So we see that is nothing but Repeat the same problem when b < 0. copyrights © youmarks.com