COMPLEX QUANTITIES 41 



Thus V -21 +16t = V - 1(21 - 16t) 



= i (4-868 l-645i } from the previous example 

 = 4-868- l-645i 2 

 = 1-645+ 4-868i 



28. The Trigonometrical Form of a Complex Quantity. When 

 we wish to raise a complex quantity to an integral or fractional 

 power, the work is simpler and more definite if we transform it 

 into its equivalent trigonometrical form. 

 Thus if a + bi = r(cos + i sin 0) 

 Equating the real and imaginary parts 



r cos = a} 

 r sin 6 = bf 

 Squaring and adding r^cos 2 -f sin 2 0) = r 2 = a 2 + b z 



Dividing tan = - 







_ 

 Hence a+bi= r(cos + i sin 0) if r = V 2 + b 2 and tan = - 







It must be remembered that this transformation is to be done 

 with the direct object of raising a complex quantity to a power 

 or extracting the root of a complex quantity. Hence if we wished 



to find the nth root of a + bi, we should have to find the nth root 



i i^ 



of r(cos + i sin 0), or the value of r(cos + i sin 0). Now if r 



i 

 is negative and n is even, we cannot find the value of r without 



introducing a further imaginary quantity, but when n is odd we 



i 



could extract the nth root. If r is positive, the value of r is real 

 whether n is odd or even. It is therefore advisable to perform 

 the above transformation with the condition that r must always 

 be positive, and so the value of the angle depends not only 

 upon the values of a and b, but also upon the signs of a and b. 

 We will next consider the different cases depending upon the 



signs of a and b. Let A be the acute angle whose tangent is -. 



Case I. When a and b are positive. 



Then r sin = b and r cos = a 



The angle must be such that its sine and cosine are both 

 positive : an angle between and 90. Thus = A. 



_ I 



Therefore a + H = Va 2 + b 2 (cos A + i sin A) where tan A = 







Case II. When a is negative and b is positive. 

 Then a + bi = r(cos + i sin 0) 

 and r cos = a and r sin = 6 



