1. ## Solving complex Equation

Q1.
Solve equa for z=a+bi where a and b are real numbers.
(z-zbar)^2=2z+6i

The ans is -18-3i but do not know how they get this.

2. $z-\overline{z}=2\text{Im}(z)i$

3. for $z=a+bi$

$(z-\bar{z})^2=2z+6i$ is

$(a+bi-(a-bi))^2=2(a+bi)+6i$

Now expand this expression out.

4. Thanks for that ,But i did that and and got -2b^2=a+bi+3i
The ans is supposed to be -18-3i

5. I get

$(a+bi-(a-bi))^2=2(a+bi)+6i$

$(a+bi-a+bi)^2=2a+2bi+6i$

$(2bi)^2=2a+2bi+6i$

$4b^2(-1)=2a+2bi+6i$

$-4b^2=2a+2bi+6i$

$-4b^2-2a-2bi=0+6i$

Now equating coefficients

$-2b=6 , -4b^2-2a=0\implies b = -3 ,-4(-3)^2-2a=0$

$\implies -4\times 9-2a=0\implies -36-2a=0, a=-18$

6. Thanks for that.
I have another one Iam having difficulty with too iff your interested.

z^2 = 3i

7. $z^2 = 3i$

$(a+bi)^2 = 0+3i$

$(a+bi)(a+bi) = 0+3i$

$a^2+2abi+(bi)^2 = 0+3i$

$a^2+2abi-b^2 = 0+3i$

$a^2-b^2 = 0, 2abi=3i$

Can you solve it form here?

8. $3i = 3\left(\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}}\right)$

$= 3e^{\frac{\pi i}{2}}$.

So $z^2 = 3i$

$z^2 = 3e^{\frac{\pi i}{2}}$

$z = \left(3e^{\frac{\pi i}{2}}\right)^{\frac{1}{2}}$

$z = \pm\sqrt{3}e^{\frac{\pi i}{4}}$

$z = \pm\sqrt{3}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$

$z = \pm\sqrt{3}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}i}{2}\right)$

$z = \frac{\sqrt{6}}{2} + \frac{\sqrt{6}i}{2}$ or $z = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}i}{2}$.

9. 'pickslides' Thats actually where I got up to and then I got stuck in finding a and b.
if you can go a bit futher would be good.

10. $a^2-b^2=0 \implies a^2=b^2 \implies a=b \implies 2b^2= 3$

11. Thanks,the ans in my text book says..... +-sqrt(3/2)*(1+i)

12. Originally Posted by pickslides
$a^2-b^2=0 \implies a^2=b^2 \implies a=b \implies 2b^2= 3$
Actually $a^2 = b^2 \implies |a| = |b|$...

13. Originally Posted by heatly
Thanks,the ans in my text book says..... +-sqrt(3/2)*(1+i)
Which is the same as what I put, I however have rationalised the denominator.

14. Thanks Prove It