Question

If $arg\left(z^{3/8}\right)=\frac{1}{2}arg(z^2+\overline{z}{z}^{1/2})$, then

• A. $\left|z\right|=1$
• B. $z=\overline{z}$
• C. $Re\left(z\right)=0$
• D. $Im\left(z\right)=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $z=\overline{z}$
B $Im\left(z\right)=0$
D $\left|z\right|=1$

A,B,D

We have,

$arg\left(z^{\frac{3}{8}}\right)=\frac{1}{2}arg(z^2+\overline{z}{z}^{\frac{1}{2}})\Rightarrow arg\left(z^{\frac{3}{4}}\right)-arg(z^2+\overline{z}{z}^{\frac{1}{2}})=0$ (since $2arg\left(z\right)=arg\left(z^2\right)$)

$\Rightarrow arg\left(\frac{z^2+\overline{z}{z}^{\frac{1}{2}}}{z^{\frac{3}{4}}}\right)=0\Rightarrow Im(z^{\frac{5}{4}}+\frac{\overline{z}}{z^{\frac{1}{4}}})=0$

$\Rightarrow z^{\frac{5}{4}}+\frac{\overline{z}}{z^{\frac{1}{4}}}=({\overline{z}}^{\frac{5}{4}}+\frac{z}{{\overline{z}}^{\frac{1}{4}}})\Rightarrow z^{\frac{5}{4}}+\frac{\overline{z}}{{\left|z\right|}^{\frac{1}{2}}}{\overline{z}}^{\frac{1}{4}}=({\overline{z}}^{\frac{5}{4}}+\frac{z{z}^{\frac{1}{4}}}{{\left|z\right|}^{\frac{1}{2}}})$

$\Rightarrow z^{\frac{5}{4}}-{\overline{z}}^{\frac{5}{4}}=\left(\frac{z^{\frac{5}{4}}-{\overline{z}}^{\frac{5}{4}}}{{\left|z\right|}^{\frac{1}{2}}}\right)\Rightarrow (z^{\frac{5}{4}}-{\overline{z}}^{\frac{5}{4}})(1-\frac{1}{{\left|z\right|}^{\frac{1}{2}}})=0$

thus, $z=\overline{z},\left|z\right|=1$