How to find polar and rectangular form of $z$ if $z^2$ is given?

Question

For example, if you have a question: $$z^2 = 3(9 + √8i + 16i^2)$$ how would you find its polar and rectangular form? the general equation is $z = a+bi$ but in this equation simplifying it would give $\sqrt{i}$, how would i go about it?

Answer 1

It is very easy to square root a complex number when it is in modulus argument form. First of all, get rid of the $i^2$ abd replace it with -1. You can then find the modulus of $z^2$ by adding the squares of the real and imaginary parts and then square rooting, i.e by using Pythagoras' theorem. You can then find the argument, $\theta$, given by $\tan{\theta}=\frac{y}{x}$, where x is the real part of $z^2$ and y is the imaginary part. To find z, square root the modulus and divide the argument by two. The other solution is this multiplied by minus one.

Answer 2

A direct attack is by

$$z^2=(x+iy)^2=x^2-y^2+2ixy=a+ib$$

equivalent to the system

$$\begin{cases}x^2-y^2&=a,\\2xy&=b.\end{cases}$$

Multiplying by $x^2$,

$$x^4-x^2y^2=x^4-\frac{b^2}4=ax^2$$ is a biquadratic equation that has two real solutions in $x,y$.

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In polar form,

$$z^2=(re^{i\theta})^2=r^2e^{2i\theta}=\sqrt{a^2+b^2}e^{i\arctan b/a}$$

gives

$$r=\sqrt[4]{a^2+b^2},\\\theta=\frac12\arctan\frac ba+ik\pi.$$

Answer 3

In polar form $$ a+ib= \sqrt{a^2+b^2}e^{i\cdot \tan^{-1} \dfrac{b}{a}}$$ For square root of a complex number we take square root of absolute value and half the modulus of $z$.

$$z^2= 3 (-9+\sqrt8 i)$$

$$ z =\sqrt{3 \, (89)}\tan^{-1} ( \frac12\,\dfrac{\sqrt8 }{-9}).$$