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As $z$ is complex, we can set $z=x+iy$ where $x,y$ are real and $i=\sqrt{-1}$ So, $z^2=(x+iy)^2=x^2-y^2+2xyi$ Equate the real & the imaginary part of $x^2-y^2+2xyi=i$

2019年10月9日 · Yes you missed something: the simplest way to solve this equation. Recall that if two complex numbers $a$ and $b$ satisfy $a^2=b^2$ then $a=b$ or $a=-b$. Now: $$z^2=-1=i^2\iff z=i\textbf{ or } z=-i$$ and it is over.

Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

If $|z^2-1|=|z|^2+1$, how do we show that $z$ lies on imaginary axis ? I understand that I can easily do this if I substitute $z=a+ib$. How do we solve it using algebra of complex numbers without ...

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)

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