$Arg(z+1) = \frac{π}{6}$ and $Arg(z1) = \frac{2π}{3}$
Solution 1
Think of it geometrically.
(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 $A=Z1$ 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 $B$,
$$\frac{\sqrt 3}{3}u=\sqrt 3(u2)$$
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
$$ \arg(z+1)=\frac\pi6\implies z+1=r_1e^{i\pi/6}\tag{1} $$ $$ \arg(z1)=\frac{2\pi}3\implies z1=r_2e^{i2\pi/3}\tag{2} $$ 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 $$ \begin{align} 1+r_1\frac{\sqrt3}2 &=1r_2\frac12\\ &=1r_1\frac1{2\sqrt3}\tag{4} \end{align} $$ which gives $r_1=\sqrt3$, and therefore, $$ z=\frac12+\frac{\sqrt3}2i\tag{5} $$
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Ivy
Updated on October 05, 2022Comments

Ivy 19 days
I'm really stuck
I need to find z when $$Arg(z+1) = \frac{π}{6}$$ and $$Arg(z1) = \frac{2π}{3}$$ Please help!!!!

robjohn over 7 yearsWhat does it mean for the argument of a complex number to be $\frac\pi6$? $\frac{2\pi}3$?


David over 3 yearsVery nice solution.

David over 3 yearsReally well done.