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Could anyone help me how to find a path in $\mathbb{C}$ plane

(1) with initial point $1-i$ and terminal point $-2+4i$ having arc length $100$ unit?

(2) How to find a smooth figure 8 path in plane?

(3) Find a path that traces $e^{2\pi it}+1$ in the opposite direction and 3 times faster.

I have simply no idea how to proceed. for the first one I though about $xt+(1-t)y$

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    For the first one, the path cannot be a straight line segment, since that would have length $\sqrt{34}=5.8309...$ So you'd have to go around a circle somehow, or along zig-zags, during the path. – 2017-01-30
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    circular path I thing.. – 2017-01-30
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    Or spiral maybe – 2017-01-30

2 Answers 2

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(1)

Let $1-i\to a-i\to a+4i\to-2+4i$ and find $a\in\mathbb{R}$.

(2)

Probably you mean Figure Eight Curve

(3)

$z=e^{2\pi it}+1$ is the circle $|z-1|=1$ or $z=1+\cos2\pi t+i\sin2\pi t$ where $0\leq t\leq1$. In this case with $1-t$ the path traced in apposite direction and the factor $\dfrac{t}{3}$ instead of $t$ caused the path spanned 3 times faster. So desired path is $z=1+\cos2\pi(1-\dfrac{t}{3})+i\sin2\pi(1-\dfrac{t}{3})$ where $0\leq t\leq3$.

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    need explaination – 2017-01-30
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    @miosaki I'm in doubt, OP says _arc_ but i assume segments. – 2017-01-30
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    Arclength of a curve is just its length as a piecewise (rectifiable) curve so this answer is OK in that sense. – 2017-01-30
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    @coffeemath Thanks. – 2017-01-30
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Hint for (1) assuming an arc of circle :

The center $O$ of a circle that passes through $A=(1,-1)$ and $B=(-2,4)$ is located on the perpendicular bissector of $[AB]$ and the arc length is $l=R\theta$ or $l=R(2\pi-\theta)$ where $\theta=\widehat{AOB}$