Question 1d

SCH_CALC L3CALC

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$$z=r\text{cis}\theta ,r>1,0<\theta <\frac{\pi }{2}$$

Convert $z$, $\overline{z}$, $\frac{1}{z}$, $iz$ in $x+yi$ form:

$$z=x+yi$$ $$\overline{z}=x-yi$$ $$\frac{1}{z}=\frac{1}{x+yi}=\frac{1}{x+yi}\cdot \frac{x-yi}{x-yi}=\frac{x-yi}{x^2+y^2}$$ $$iz=i(x+yi)=-y+xi$$

Plotting the coordinates on the Argand diagram, where $\angle ABC=60^o$ and $BC=1.5$:

To get the area of ABCD, looks like I can sum the area of ABO and ADO, then subtract that sum by $CDO$.

$$A_{ABCD}=(A_{ABO}+A_{ADO})-A_{CDO}$$

Get $A_{ABO}$:

$$A=\frac{1}{2}ab\sin C$$

Let $CO=x⟹OB=x+1.5$

$$A=\frac{1}{2}(x+1.5{)}^2\cdot \sin \frac{\pi }{3}$$ $$A_{ABO}=\frac{\sqrt{3}}{4}(x+1.5{)}^2$$

Now get $A_{ADO}$:

$$A=\frac{1}{2}{\left(x+1.5\right)}^2\cdot \sin \frac{\pi }{2}$$ $$A_{ADO}=\frac{1}{2}(x+1.5{)}^2$$

Then, get $A_{CDO}$:

$$A=\frac{1}{2}ab\cdot \sin C$$ $$A=\frac{1}{2}(1.5+x)(x)\cdot \sin C$$

Looking at the diagram, I can see that $\angle C=90^o+30^o+30^o=150^o\equiv \frac{5\pi }{6}$

$$⟹C=\frac{5\pi }{6}$$

Hence

$$A=\frac{x}{2}(x+1.5)\cdot \frac{1}{2}$$ $$A_{CDO}=\frac{x}{4}(x+1.5)$$

Now obtaining a formula for $A_{ABCD}$:

$$A_{ABCD}=\left[\frac{\sqrt{3}}{4}(x+1.5{)}^2+\frac{1}{2}(x+1.5{)}^2\right]-\left[\frac{x}{4}(x+1.5)\right]$$

After simplification:

$$A=\frac{1+\sqrt{3}}{4}{x}^2+\frac{9+6\sqrt{3}}{8}x+\frac{18+9\sqrt{3}}{16}$$

Now we need $x$.

Let $x+1.5=r$

$$⟹r-\frac{1}{r}=1.5$$ $$⟹r^2-1.5r-1=0$$ $$⟹r=2,-\frac{1}{2}$$

But $r>1$, so $r=2$

$$r=2⟹x+1.5=2⟹x=0.5$$

Now substitute $x=0.5$ into the area formula:

$$A=\frac{1+\sqrt{3}}{4}(0.5{)}^2+\frac{9+6\sqrt{3}}{8}x+\frac{18+9\sqrt{3}}{16}$$ $$\boxed{A_{ABCD}=\frac{7+4\sqrt{3}}{4}}$$