STEP II 1997 question 5 solution

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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 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 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 with the positive x-axis.