Finding the Complex Solution to a Given Equation?

Question

How do we find a complex number(s) $z = a+bi$ satisfying $z^2 = \frac{1}{2}(a-bi)$

I don't know how to do this. Any hint is appreciated.

Answer 1

Hint: multiply both sides by $z$ to discover that $z$ must be the cube root of a positive real number (which gives three possibilities for its argument). Then take the modulus of both sides to discover its modulus. This narrows down to only three (non-zero) possibilities. Check them individually to find which are genuine solutions.

Answer 2

Hint: We need: $ 2(a + bi)^2 = a - bi $. Simplifying, $ (a + bi)^2 = a^2 - b^2 + 2abi $.

Hover over these blocks to see the next steps:

Then equating the real and imaginary parts we have, $ a = 2(a^2 - b^2) $ and $b = -4ab$.

Then either:
1. $a = b = 0$, or
2. $a = \frac{1}{2}$ and $b = 0$, or
3. $a = -\frac{1}{4}$ and $b = \sqrt{\frac{1}{16} + \frac{1}{8}} = \sqrt{\frac{3}{16}} = \pm\frac{\sqrt{3}}{4}$

$$ z = -\frac{1}{4} \pm i\frac{\sqrt{3}}{4} \qquad\text{or}\qquad z = \frac{1}{2} \qquad\text{or}\qquad z = 0 $$

Answer 3

Substitute your definition for $z$ into the equation:

$$ \begin{align} \frac{a-ib}{2}&=(a+ib)^2\\ &=a^2+2iab-b^2 \end{align} $$ giving us $$ \frac{a}{2}=a^2-b^2 $$ and $$ \frac{-ib}{2}=2iab $$ cancelling terms on both sides in this second equation, we get $$ -\frac{1}{4}=a $$ and substituting this value into the first equation we have $$-\frac{1}{8}=\frac{1}{16}-b^2$$ which gives $$b=\pm\frac{\sqrt{3}}{4}$$ In this case, you must also test $a=0$ and $b=0$. When $b=0$ $$\frac{a}{2}=a^2\\a=\frac{1}{2}$$ when $a=0$ $$\frac{-ib}{2}=-b^2$$ which has an imaginary solution. $b$ must be real, so there is no solution. We have found $3$ solutions.

Answer 4

HINT: Starting with $z = a + bi$, find $z^{2}$ and solve the given equation by setting the real part of one side equal to the real part of the other; similarly for the imaginary parts. You should then be able to solve explicitly for $a$; upon which, you will re-substitute that value to get an equation involving only $b$. You will then solve for $b$---I think there will be multiple $b$'s. Then, you can construct the solutions.

Answer 5

Let $z=re^{it}$ where $r\ge0,t$ are real

We have $$2(re^{it})^2=re^{-it}$$

Take modulus, $$2r^2=r\implies r=?$$

If $2r=1,$

$$e^{3it}=\dfrac1{2r}=?=e^{2m\pi i}$$

$3it=2m\pi i$ where $m$ is any integer

$t=\dfrac{2m\pi}3$ where $m=-1,0,1$