8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.

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1 8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the mnner in which we lnce our checkooks or count the cns of sod wter in the pntry. But, if complex numers hd never een discovered, the development of modern coustic equipment nd highly integrted circuit ords would hve een either delyed or perhps impossile. In the simplest setting, we see the first need for dditionl numers when we notice tht we cn not solve the eqution x + = 0 using rel numers. This is ecuse when we squre rel numer we get nonnegtive numer. Consequently, when we dd to the result, we get positive numer. Since we cn never hve positive numer equl to 0, we cn not find rel numer solution to x + = 0. Now, if we hd discovered (fter hundreds of yers of scientific explortion) tht we must hve solution to this eqution, then we might simply define new numer tht gives us the solution. Noting tht x + = 0 is equivlent to x =, we re led to the following definition. Definition 8.: We define i to e the imginry numer equl to the squre root of -. Tht is i =, which implies i =. We cn comne the rel numers with this new imginry numer to form the complex numers. Definition 8.: The set of complex numers is given y = { +, } The opertions of ddition, sutrction nd multipliction re defined on s follows: + + c+ di = + c + + d i for every cd,,,. ( ) ( ) ( ) ( ) ( + ) ( c+ di) = ( c) + ( d) i for every cd,,,. ( + )( c + di) = ( c d ) + ( d + c) i for every cd,,,. In ddition, the conjugte of complex numer is given y + = for every,. mrk 8.3: The set of complex numers contins the set of rel numers since = + 0i for every. Exmple 8.4: ( )( ) ( )( ) ( ) ( ) i + + = + = = +. So, ( + 3i)( + 3i) = + 3 = 3 nd ( i)( i) ( ) 4 4 = 4 + = 0. The conjugte of complex numer cn e used to develop formul for the division

2 of two complex numers. Formlly, we expect = = = i Similrly, we expect in generl, c+ di c+ di ( c + d) + ( d c) i c+ d d c = = = i Definition 8.5: Division is defined on vi c+ di c+ di ( c + d) + ( d c) i c+ d d c = = = i for ll cd,,, with + 0. It is possile to give geometric interprettion of. First, note tht every numer in hs the form + where,. The rel prt of this complex numer is nd the imginry prt of this complex numer is. Tht is, the complex numer + is completely determined y the two vlues nd. As result, it is tempting to give geometric interprettion of + y viewing the point (, ) in coordinte system. When this is done, the horizontl xis is referred to s the rel xis nd mrked, the verticl xis is referred to s the imginry xis nd mrked, nd the entire coordinte system is referred to s the complex plne. We show the complex numer + grphed elow in the complex plne long with its conjugte + =. The Complex Plne + Notice tht the conjugte of complex numer cn e viewed geometriclly s the reflection of the complex numer cross the imginry xis. We cn lso use the picture ove to define the notion of solute vlue of complex numer. cll tht the solute vlue of rel numer is its distnce from 0. We cn extend this to the setting of complex numers y noting tht 0 cn e written in s 0+ 0i. So 0 is corresponds with the origin in the complex plne, nd we should require + to e

3 the distnce from the point + in the complex plne to the origin. We cn find this vlue using the Pythgoren theorem s shown elow. + + This motivtes the following definition. Definition 8.6: If, then the solute vlue of + is given y Exmple 8.7: Compute 3 + i. + = + Solution: 3+ i = 3 + = 0 Exmple 8.8: Give geometric interprettion of ll complex numers + stisfying + =. Solution: The complex numers stisfying + = correspond to the points in the complex plne tht re unit wy from the origin. This is exctly the circle of rdius centered t the origin. Every point on the circle in the grph elow stisfies + =. Notice tht this includes the rel vlues - nd long the rel xis, which re the rel numers tht stisfy x =.

4 + Exercises. Compute ( 3 i) 3( 4 i) ( i)( i). Compute ( 3 i) ( i)( + i) Write i in the form + for some,. 3+ i 4. Show tht if, then + =. Give geometric interprettion of this result. 5. Give geometric interprettion of complex ddition. Hint: Plot rtrry points + nd c+ di in the complex plne. Then drw line segments from the origin to ech of these points nd complete prllelogrm s shown in the figure. c+ di + Show tht the complex numer ( + ) + ( c+ di) is the dditionl point shown in the figure. 6. Use the ides from the exercise ove to give geometric interprettion of sutrction of complex numers. 7. Give geometric description of the set of complex numers + tht stisfy

5 + =. 8. Give geometric description of the set of complex numers + tht stisfy ( + ) ( i) =. 9. Show tht the set of complex numers is n uncountle set. 0. There is no nturl wy to define < nd > in the complex plne so tht every pir of complex numers + nd c+ di cn e compred. However, it is possile to define < nd > for some pirs of complex numers. To this end, we sy tht + < c+ di if nd only if < c nd < d for cd,,,. Drw picture in the complex plne of the set of complex numers + stisfying + < i.. Use the exercise ove to give definition of + c+ di for cd,,,. Then sketch the set of complex numers + stisfying + 3i +.. Give geometric interprettion of the set of complex numers stisfying Give geometric interprettion of the set of complex numers + stisfying + nd 0 < Give geometric interprettion of the set of complex numers + stisfying + > nd Show tht the opertions ddition nd multipliction stisfy the field xioms on. Tht is, for every c,,. Nme Addition Multipliction Commuttive + = + = Associtive ( ) c ( c) Distriutive Identity 0 0 Inverse There exists so tht + = + = = + + ( ) c = ( c) ( + c) = + c + = + = ( ) ( ) ( ) ( ) = = if 0 there exists so tht = =