Trigonometric Form of Complex Numbers If z = a + bi = r (cos + i sin ) = rcis , then: z = a bi = r (cos i sin ), but "cis" form is cos + i sin , i.e., a PLUS sign between the sine and cosine. Now, one can argue that the PLUS/MINUS doesn’t really matter, and that’s true to an extent, but I need proof that YOU understand why that is the case ... and here is WHY: cos ( ) = cos and sin ( ) = sin , so r (cos i sin ) = r (cos ( ) + i sin ( )) = rcis ( ) . (note that both sine and cosine are of the same angle ) So, this means z = a bi = r (cos i sin ) = rcis( ) : Furthermore, if you plot the conjugate of z, you realize that its argument (i.e., angle) is the same, but in the opposite direction of z. Consider 1=z. We know that 1 1 1 = = z a + bi a + bi The numerator is the conjugate of z, i.e., z = a bi, and a2 + b2 = r2 . From above we know that a bi = rcis( 1 z = = = a bi a bi = 2 . a bi a + b2 the denominator is the square of the magnitude, i.e., ) : So we have a bi a2 + b2 rcis ( ) r2 1 cis ( ) . r – Note that 1 1 = ; z rcis although true, is not in "trigonometric form" because it is equivalent to 1 1 = ; rcis r (cos + i sin ) which is not quite in the form R (cos a sine of the same angle. + i sin ). We can’t "split the denominator" and get a cosine and – However - we can convert the expression to R (cos "conjugate": 1 rcis = = = = + i sin ) form easily enough by multiplying by the 1 r cos r (cos + i sin ) r cos r cos ri sin r2 cos2 + r2 sin2 r (cos i sin ) r2 cos2 + sin2 1 cis ( ) . r ri sin ri sin – Alternatively ... note that 1 1cis0 = = (1 r) cis (0 z rcis ) = rcis ( ). Wow. That was a LOT easier. If you plot z you will see that it is simply a rotation of z through an angle of 180 ; or radians. Hence z = rcis( + ). Again, although rcis is equivalent, we’re not always convinced students know what’s happening in this case. Is a student "making r negative," or are they "taking the opposite of r"? Use DeMoivre’s Theorem to determine zi. Note that zi = (rcis ) 1cis 2 = rcis + 2 . In words, multiplying a complex number by i amounts to a counter-clockwise rotation by 90 degrees or =2 radians. The notation jzj ; as always, represents the magnitude of the complex number z, which is simply r. p Express 34cis tan 1 53 in rectangular form (without using a calculator). Draw a sketch, keeping in mind that if tan = 35 , then xy = 35 . So the ratio of the y-coordinate to the x-coordinate must be -5:3 (and it just so happens that this produces the correct hypotenuse as well). Without our calculators, we cannot determine the angle, but we can still draw the triangle! From the above triangle we can …nd the parts we need. Note that = tan and from the diagram that cos = p p3 34 and sin = 34cis tan 1 5 3 1 p5 , 34 5 3 , so the trig form becomes = p 34 (cos + i sin ) = p 34 = 3 3 5 p + ip 34 34 5i. – Uh ... Wait a minute ... there are TWO angles that have a tangent of 35 . How did we know to choose the 4th quadrant? Here is how: we didn’t really HAVE a choice because the inverse tangent function only produces angles in the range 2 ; 2 . This removes the 2nd quadrant option from consideration! – Uh ... Wait another minute ... look at the diagram - isn’t the complex number 3 5i pretty obvious from the picture? Perhaps we should have simply recognized that the information given should have been enough without going through all the number-crunching!
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