Question

If a complex number z and $z+\frac{1}{z}$ have same argument then-

• A. z must be purely real
• B. z must be purely imaginary
• C. z cannot be imaginary
• D. z must be raal

Answer 1

The correct option is B z must be purely imaginary

we have,

z is a complex number.

Then,

Let $z=r\cos \theta +i\sin \theta$

Now, we know that,

$\overline{z}=r(\cos \theta -i\sin \theta )$

So,

$z\overline{z}=r^2$

Then,

$\frac{1}{z}=\frac{\overline{z}}{z\overline{z}}=\frac{\overline{z}}{r^2}$

Now,

According to given question,

$z+\frac{1}{z}=|z+\frac{1}{z}|=1\left(\text{z}\text{and}\text{z+}\frac{\text{1}}{\text{z}}\text{same argument}\right)$

$=|r(\cos \theta +i\sin \theta )+\frac{r(\cos \theta -i\sin \theta )}{r^2}|=1$

$=|(r+\frac{1}{r})\cos \theta +i(r-\frac{1}{r})\sin \theta |=1$

Now, squaring both side and we get,

$|(r^2+\frac{1}{r^2}+2){\cos }^2\theta +i^2(r^2+\frac{1}{r^2}-2){\sin }^2\theta |=1$

So,

$(r^2+\frac{1}{r^2}+2){\cos }^2\theta =1$

$r^2+\frac{1}{r^2}+2\cos 2\theta =1$

$r^2+\frac{1}{r^2}=1-2\cos 2\theta$

Now,

$1-2\cos 2\theta =0$

$\Rightarrow \cos 2\theta =\frac{1}{2}$

$\Rightarrow \cos 2\theta =cos\frac{\pi }{3}$

$\Rightarrow 2\theta =\frac{\pi }{3}$

$\Rightarrow \theta =\frac{\pi }{6}$

Then, Z is purely imaginary.

Hence, this is the answer.