# Question Video: Finding the Image of Complex Functions under a Given Transformation

A transformation which maps the π§-plane to the π€-plane is defined by π : π§ βΌ 1/π§, where π§ β  0. 1) Find an equation for the image of |π§| = 2 under the transformation. 2) Find an equation for the image of the arg(π§) = 3π/4. 3) Find a Cartesian equation for the image of Im(π§) = 2. 4) Find a Cartesian equation for the image of |π§ β π| = 1.

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### Video Transcript

A transformation which maps the π§-plane to the π€-plane is defined by π taking π§ to one over π§, where π§ is nonzero. Part one, find an equation for the image of the modulus of π§ equals two under the transformation. Part two, find an equation for the image of the argument of π§ equals three π by four. Part three, find a Cartesian equation for the image of the imaginary part of π§ equals two. And part four, find a Cartesian equation for the image of the modulus of π§ minus π equals one.

Our transformation takes π§ to its reciprocal, one over π§. And so, π€ is one over π§. In the first part, we want to find the image of the modulus of π§ equals two under this transformation. We can invert our transformation to find π§ in terms of π€. Multiplying both sides by π§ and then dividing through by π€, we find that π§ is one over π€. Substituting this then, we find that our image has the equation the modulus of one over π€ equals two. But we can improve on this equation. We use the fact that the modulus of a quotient is the quotient of the moduli and that the modulus of one is simply one. Rearranging then, we can rewrite our equation as the modulus of π€ equals a half.

Now what does this look like on the complex plane? Well, our original locus with equation the modulus of π§ equals two is a circle with centre the origin and radius two in the π§-plane. And its image under the reciprocal transformation, weβve shown has equation the modulus of π€ equals a half, which we recognise as a circle with centre the origin and radius a half in the π€-plane. This makes sense. We know that the image of a complex number with modulus π under this transformation will have a modulus one over π. And so the image of a complex number with modulus two has modulus one over two which is a half. And so, when we map the circle of all complex numbers with modulus two under this transformation, itβs natural that we get the circle of all complex numbers with modulus a half.

Letβs now move on to find the image of the argument of π§ equals three π by four. Again, we use the fact that π§ is one over π€. And so the argument of one over π€ is three π over four. We use the fact the argument of a quotient is the difference of the arguments. And also that the argument of one is just zero. And so multiplying both sides by negative one, we get the equation the argument of π€ equals negative three π by four. Again, itβs helpful to look at the diagram. We see that the half line of complex numbers with argument three π by four is mapped to the half line complex numbers with arguments negative three π by four. We know thereβs a complex number with argument π will be mapped to a complex number with argument negative π. So this makes sense.

Now letβs find the image of the imaginary part of π§ equals two. It may not be clear what to do with the imaginary part of one over π€. What we do as we ask for a Cartesian equation is to write π€ in terms of its real and imaginary parts, which we call π’ and π£. How do we find the imaginary part of one over π’ plus ππ£? We make the denominator real in the normal way. And now with this number written in algebraic form, we can just read off the imaginary parts. Itβs negative π£ over π’ squared plus π£ squared. We might be tempted to stop simplifying at this point. But if we divide by two and complete the square in π£, we get something. This is recognisably the equation of a circle. Letβs plot this on a diagram. We see that the line of complex numbers with imaginary part two has been mapped to a circle in the π€-plane.

Finally, we find the image of the modulus of π§ minus π equals one. We substitute one over π€ for π§ and then write what we have inside the modulus as a single fraction. This allows us to apply what we know about the modulus of a quotient. So we can multiply both sides by the modulus of π€. Now we can substitute π’ plus ππ£ for π€. We can simplify on the left-hand side. And now, weβre ready to apply the definition of the modulus. We square both sides. And negative π’ squared is the same as π’ squared; so these cancel. Distributing on the left-hand side, we see the π£-squared terms cancel as well. So we find that two π£ is negative one. And hence, the equation of our image is π£ equals negative a half. We see then that the circle with centre π and radius one is mapped to the straight line of points in the π€-plane with imaginary part negative a half.

Letβs recap then. In part one, the circle was mapped to a circle. In part two, a line was mapped to a line. Well, actually, a half line was mapped to a half line. In part three, a line was mapped to a circle. And in part four, a circle was mapped to a line.