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STEP II 1997 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP II 1997 question 5 solution


If z = ie, then if we equate real and imaginary parts of the equation z = e^w we get:

e^u  \cos (v) = 0

e^u  \sin (v) = e


Since e^u > 0, then  \cos (v) = 0 and thus v = \pi/2 + k\pi. Whence e^u = e, and so u=1. That is,

w = 1 + i(\frac \pi 2 + k\pi)


If u is constant, say u = c, then

x + iy = e^c  \cos (v) + i e^c  \sin (v)

\displaystyle \Longrightarrow \frac{x}{e^c} =  \cos (v) \text{  and  } \frac{y}{e^c} =  \sin (v)


Thus

x^2 + y^2 = e^{2c}


So the corresponding locus in the x-y plane is simply a circle centered at the origin of radius e^c.


On the other hand, if v is constant, say v=k, then

x + iy = e^u  \cos (k) + i e^(u)  \sin (k)

\displaystyle \Longrightarrow \frac{x}{ \cos (k)} = e^u = \frac{y}{ \sin (k)}


Thus

y =  \tan (k)\, x


That is, the corresponding locus in the x-y plane is a straight line through the origin of gradient  \tan (k), i.e. it makes an angle of k with the positive x-axis.


So, for the last bit, the following subsets do the trick:

\displaystyle W_1 = \left\{ (u,v) : u \in \mathbb R, 0 < v < \frac \pi 2 \right\}

\displaystyle W_2 = \left\{ (u,v) : u \in \mathbb R, 2\pi < v < \frac{5\pi}{2 }\right\}

\displaystyle W_3 = \left\{ (u,v) : u < 0, 0 \leq v < 2\pi \right\}

\displaystyle W_4 = \left\{ (u,v) : u < 0, 2\pi <= v < 4\pi \right\}

Solution by dvs

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