$Arg(z+1) = \frac{π}{6}$ and $Arg(z-1) = \frac{2π}{3}$

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Solution 1

Think of it geometrically.

enter image description here (The angles are shown in degrees rather than radians.) You want point $B=Z+1$ on the ray with argument $\frac{\pi}6=30°$ and point $A=Z-1$ on the ray with argument $\frac{2\pi}3=120°$. In other words, you want to find the horizontal line such that the distance between the intersection points is $2$.

As you can see, the triangle $BAO$ is a right triangle, with hypotenuse $2$. You should be able to find the altitude of that triangle, which will be the imaginary part of $Z$. The real part of $Z$ should then be easy to find. Or, you could find the equations of rays $OB$ and $OA$ and find the value of $y$ where the $x$ values differ by $2$.

Doing this the geometric way, we see that triangle $BAO$ is a $30°$-$60°$-$90°$ triangle with hypotenuse $2$. Therefore the legs are $1$ and $\sqrt 3$, the area is $\frac{\sqrt 3}2$, and the altitude is also $\frac{\sqrt 3}2$. Thus the $y$-coordinate of point $B$ is $\frac{\sqrt 3}2$, the $x$-coordinate is $\frac 32$, and the $x$-coordinate of point $Z$ is $\frac 12$. Putting this together for point $Z$ we get

$$z=\frac 12+\frac{\sqrt 3}{2}i$$

Doing this my second way, the equation of line $OB$ is $y=\frac{\sqrt 3}{3}x$ (since $\tan\frac{\pi}{6}=\frac{\sqrt 3}{3}$), and the equation of line $OA$ is similarly $y=-\sqrt 3x$. Letting $u$ be the $x$-coordinate of point $B$,

$$\frac{\sqrt 3}{3}u=-\sqrt 3(u-2)$$

Solving this gives $u=\frac 32$, so point $B$ is $(\frac 32,\frac{\sqrt 3}{2})$. This means point $Z$ is $(\frac 12,\frac{\sqrt 3}{2})$.

Therefore,

$$z=\frac 12+\frac{\sqrt 3}{2}i$$

This answer checks. (The graphic was updated to place the points in the correct places, reflecting the final answer.)

Solution 2

$$ \arg(z+1)=\frac\pi6\implies z+1=r_1e^{i\pi/6}\tag{1} $$ $$ \arg(z-1)=\frac{2\pi}3\implies z-1=r_2e^{i2\pi/3}\tag{2} $$ for some $r_1,r_2\ge0$.

Putting $(1)$ and $(2)$ together yields $$ z=-1+r_1\left(\frac{\sqrt3}2+\frac12i\right)=1+r_2\left(-\frac12+\frac{\sqrt3}2i\right)\tag{3} $$ For $(3)$ to be true, the imaginary parts must be equal and therefore $\frac12r_1=\frac{\sqrt3}2r_2$. That is, $r_1=\sqrt3\,r_2$. Equating the real parts, we get $$ \begin{align} -1+r_1\frac{\sqrt3}2 &=1-r_2\frac12\\ &=1-r_1\frac1{2\sqrt3}\tag{4} \end{align} $$ which gives $r_1=\sqrt3$, and therefore, $$ z=\frac12+\frac{\sqrt3}2i\tag{5} $$

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Updated on October 05, 2022

Comments

  • Ivy
    Ivy 19 days

    I'm really stuck

    I need to find z when $$Arg(z+1) = \frac{π}{6}$$ and $$Arg(z-1) = \frac{2π}{3}$$ Please help!!!!

    • robjohn
      robjohn over 7 years
      What does it mean for the argument of a complex number to be $\frac\pi6$? $\frac{2\pi}3$?
  • David
    David over 3 years
    Very nice solution.
  • David
    David over 3 years
    Really well done.