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- 1. 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
- 2. Problem <ul><li>Find z = </li></ul><ul><li>One way to find this is by equation z to x + iy, on squaring which gives </li></ul><ul><li>7 + 24i = x 2 – y 2 + 2ixy </li></ul><ul><li>x 2 – y 2 = 7 and xy = 12 which on solving gives the value of x and y. </li></ul><ul><li>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. </li></ul><ul><li>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. </li></ul>copyrights © youmarks.com
- 3. Finding <ul><li>Again as we did, one method is put x + iy = </li></ul><ul><li>This would as before give us </li></ul><ul><li>x 2 – y 2 =a and xy = b/2. </li></ul><ul><li>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. </li></ul>copyrights © youmarks.com
- 4. Finding <ul><li>We know that </li></ul><ul><li>( c + id ) 2 = c 2 – d 2 + 2icd. To find </li></ul><ul><li>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. </li></ul><ul><li>Let us consider the case when b > 0. Now </li></ul><ul><li>b = √b 2 </li></ul><ul><li>Or </li></ul>copyrights © youmarks.com
- 5. Finding Also note that a = copyrights © youmarks.com
- 6. Finding So we see that complex number a + ib is nothing but copyrights © youmarks.com
- 7. Finding So we see a + bi is is denoted by |z| if z = a + ib copyrights © youmarks.com
- 8. Finding So we see that is nothing but Repeat the same problem when b < 0. copyrights © youmarks.com

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