Saturday, March 2, 2013

Complex Numbers Solutions (1, 2) - Polar Form

The following are the solutions for the problems appearing at the end of the Feb. 28 blog:

(1) We have:


x3 + iy3 = - 4 – i

and: arg(z3) = arctan(y3/x3) = arctan(-1/-4) = arctan(0.25)

so q = arg(z3) =  14 deg.

The value of r3= [x32 + y32]1/2 =  [(-4)2 + (-1)2]1/2 = [17]1/2 = 4.1

So: z3 = 4.1 cos(14) + isin(14) = 4.1 cis(14)

(2) Recall: C = z4 = 4 + 3i


Thence: z5 = [z3 + z4] = [(-4 – i) + (4 – 3i)] = 0 – 4i

arg(z5) = arctan(y5/x5) = arctan(4 /0) = ¥

(No polar form applicable for an infinity)  Cont'd

 

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