If the roots of the equation z^2-i=0 are alph | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation $z^2-i=0$,we have $z^2=i$. Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$.

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) Given the equation $z^2-i=0$,we have $z^2=i$. Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$. The roots are $z = \pm e^{i\frac{\pi}{4}}$. Thus,the roots are $z_1 = e^{i\frac{\pi}{4}}$ and $z_2 = e^{i(\frac{\pi}{4} + \pi)} = e^{i\frac{5\pi}{4}}$. Let $\alpha = e^{i\frac{\pi}{4}}$ and $\beta = e^{i\frac{5\pi}{4}}$. Then $\operatorname{Arg} \alpha = \frac{\pi}{4}$ and $\operatorname{Arg} \beta = \frac{5\pi}{4}$. Therefore,$|\operatorname{Arg} \beta - \operatorname{Arg} \alpha| = |\frac{5\pi}{4} - \frac{\pi}{4}| = |\pi| = \pi$.

If the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$,then $|\operatorname{Arg} \beta-\operatorname{Arg} \alpha|=$

Similar Questions

Find the modulus and the argument of the complex number $z = -1 - i \sqrt{3}$.

The amplitude of ${e^{{e^{ - i\theta }}}}$ is equal to

If $5 + ix^3y^2$ and $x^3 + y^2 + 6i$ are conjugate complex numbers and $\arg(x + iy) = \theta$,then $\tan^2 \theta$ is equal to

The value of $\frac{4(\cos 75^o + i\sin 75^o)}{0.4(\cos 30^o + i\sin 30^o)}$ is

Convert the given complex number in polar form: $-1+i$

Vedclass Test Series

Exam Paper Generator

Online Exam Module