Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2 \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$ which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$ and after that i can't get the expansion to work.
Can you help me out?
Starting from what you ought to have:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (2+\frac{3}{2}i)^2$.
we get:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (4+6i-\frac{9}{4}) = \frac{-21+20i}{4} = (\frac{2+5i}{2})^2$.
The last step I obtained by guessing. If you want a systematic way to find the square root of a complex number if it is a perfect square (the square of a rational complex number) you can use the following technique.
Given $(a+bi)^2 = c+di$ where $a,b,c,d \in \mathbb{R}$:
$a^2-b^2 = c$ and $2ab = d$ [by comparing real and imaginary parts].
Thus $4 a^2 - 4 a^2 b^2 = 4 a^2 c^2$ and $4 a^2 b^2 = d^2$ [multiply the first by $2a^2$ and square the second].
Thus $4 a^4 - 4c a^2 - d^2 = 0$ [Add them together to get a quadratic in $a$].
Thus $(2a^2-c)^2 = c^2+d^2$ [Complete the square].
Thus $2a^2-c = \sqrt{c^2+d^2}$ and hence $a = \pm \sqrt{\frac{\sqrt{c^2+d^2}+c}{2}}$.
Also $2b^2 = 2a^2 - 2c = \sqrt{c^2+d^2}-c$ and hence $b = \pm \sqrt{\frac{\sqrt{c^2+d^2}-c}{2}}$.
[Note that the choices of sign are dependent, so it may be better to use $b = \frac{d}{2a}$ instead.]
In the above case we get:
$\sqrt{-21+20i} = \pm \left( \sqrt{\frac{\sqrt{21^2+20^2}-21}{2}} + \sqrt{\frac{\sqrt{21^2+20^2}+21}{2}} i \right) = \pm (2+5i)$.
[Here I have chosen the signs correctly.]
Notice, $$(4-3i)z^2-25z+31-17i=0$$ Solving the above quadratic equation for $z$ as follows $$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$ $$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$ $$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{2(16+9)}$$ $$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{50}$$ $$z=\frac{(4+3i)\left(25\pm\sqrt{725}\left(\cos \frac{\alpha}{2}+i\sin\frac{\alpha}{2}\right)\right)}{50}$$ Where, $\alpha=\cos^{-1}\left(\frac{333}{725}\right)=\sin^{-1}\left(\frac{644}{725}\right)$
I hope you can take it from here.
Start from what you've got
$z^2 -(4+3i)z + 7+i=0$
So
$\Delta=(4+3i)^2-4(7+i)= -21+20i=(2+5i)^2$
Then you can use the formula for the solution of a quadratic equation to get
$z=3+4i\quad $ and $\quad z=1-i$.
$$(4-3i)z^2-25z+31-17i=0\Longleftrightarrow$$ $$(4-3i)((7+i)-(4+3i)z+z^2)=0\Longleftrightarrow$$ $$(4-3i)((7+i)+(-4-3i)z+z^2)=0\Longleftrightarrow$$ $$(7+i)+(-4-3i)z+z^2=0\Longleftrightarrow$$ $$(z+(-3-4i))((-1+i)+z)=0\Longleftrightarrow$$ $$z+(-3-4i)=0 \vee (-1+i)+z=0\Longleftrightarrow$$ $$z=3+4i \vee (-1+i)+z=0\Longleftrightarrow$$ $$z=3+4i \vee z=1-i$$
If $z^2−(4+3i)z+7+i=0$
Then
$\Delta=(4+3i)^2−4(7+i)=−21+20i=(2+5i)^2$.
From which one can use the quadratic formula to obtain;
$z=3+4i $ and $z=1−i ...$
