Understanding Engineering Mathematics

5.(i) √ 2 (ii) √ 5 (iii) √ 5(iv) √ 14 (v) 5 (vi) 2 6.(i) √ 14 (ii) 2 √ 3 (iii) √ 35 (iv) √ 22 (v)a √ 6 60 ° (i) ( 1 √ 2 , 1 √ ...

(i) − 5 i− 5 j+ 7 k, 1 3 √ 11 (− 5 i− 5 j+ 7 k) (ii) 2i+ 31 j+ 13 k, 1 √ 1134 ( 2 i+ 31 j+ 13 k) (iii) 7i+ 11 j− 8 k, 1 √ 234 ...

12 Complex Numbers Complex numbers extend the notion of ordinary ‘real numbers’ to include a new kind of ‘number’,j= √ −1. Such ...

12.1 What are complex numbers? Numbers such as integers, rational numbers, etc. that we have been using so far are called real n ...

Any number of this form z=a+jb j= √ − 1 wherea,bare real numbers, is called acomplex number. We will usually writezrather thanxf ...

Problem 12.3 Multiply (3−j)and(2Y 3 j). ( 3 −j)( 2 + 3 j)= 6 + 9 j− 2 j− 3 j^2 = 6 + 7 j− 3 (− 1 )= 9 + 7 j Given a complex numb ...

Division (or rationalization)of two complex numbers: z= a+jb c+jd means converting into the standard formA+jB. We can do this by ...

y y O x P r z = x + iy = r cos q + jr sin q x q Figure 12.1The Argand diagram. cisθ is shorthand for cosθ+isinθ, using the alter ...

Answers − 3 + 2 j j 1 0 2 +j 1 −j − 3 j − 3 − 2 j y x (i) 1^ ( 0 ) (ii) 1^ (π 2 ) (iii) 3^ ( − π 2 ) (iv) √ 2 ( − π 4 ) (v) ...

We have (^) (θ 1 ) (^) (θ 2 )=(cosθ 1 +jsinθ 1 )(cosθ 2 +jsinθ 2 ) =cosθ 1 cosθ 2 −sinθ 1 sinθ 2 +j(sinθ 1 cosθ 2 +sinθ 2 cosθ 1 ...

So in polar form the product becomes ( √ 3 −j)( 1 − √ 3 j)= 2 ( − π 6 ) 2 ( − π 3 ) = 4 ( − π 6 − π 3 ) = 4 ( − π 2 ) =− 4 j ...

12.5 Division in polar form You can probably guess what happens with division in polar form now – divide modulii and subtract ar ...

which we can check directly √ 3 −j 1 − √ 3 j = ( √ 3 −j)( 1 + √ 3 j) 12 +( √ 3 )^2 = 2 √ 3 + 2 j 4 = √ 3 2 + 1 2 j Exercise on 1 ...

Problem 12.11 Put into exponential form (i) 1 (ii)−1 (iii)− 2 j (iv) 1Y √ 3 j (v) √ 3 −j Conversion to exponential form is of co ...

( √ 3 +j)−^3 = ( 2 ( cos π 6 +jsin π 6 ))− 3 = 1 8 ( cos π 6 +jsin π 6 )− 3 = 1 8 ( cos ( − π 2 ) +jsin ( − π 2 )) =− 1 8 j Exer ...

but two of them are complex. The question is, how do we find such roots in general? Since the polar form is so useful in multipl ...

Problem 12.13 Evaluate ( 2 kp 3 ) fork=0, 1, 2, 3. Substituting forkwe find: k= 0 ,^ ( 2 kπ 3 ) =^ ( 0 )= 1 k= 1 ,^ ( 2 π 3 ) = ...

Exercises on 12.8 Find ina+jbform and plot on Argand diagram: (i) The three values ofj^1 /^3 (ii) The four values of( 1 +j)^1 / ...

(i) ± 1 1 2 ± √ 3 2 j − 1 2 ± √ 3 2 j (ii) − 21 ± √ 3 j 12.9 Reinforcement 1.Write the following in simplest form in terms of ...

6.Put intoa+jbform (i) ( 2 + 5 j)( 4 − 3 j) (ii) ( 4 −j)( 1 +j)( 3 + 4 j) (iii) 1 3 − 4 j (iv) 2 − 5 j 1 + 4 j (v) − 1 + 3 j ( 3 ...