MHB Complex numbers simultaneous equations

[lemonthree]

Hi all, I have spent a couple of hours on this perplexing question.

Solve the simultaneous equations:
z = w + 3i + 2 and z² - iw + 5 - 2i = 0
giving z and w in the form (x + yi) where x and y are real.

I tried various methods, all to no avail.
I have substituted z into z² , I got the wrong answers.
I also tried letting z be (a + bi) and w be (c + di) and tried to combine the 2 equations together, and I got a horrible mess with many unknowns.

Please help me, thank you!

 

[evinda]

Just substitute $z=w+3i+2$ in $z^2-iw+5-2i=0$ and use the fact that $(a+b+c)^2=a^2+2ab+2ac+b^2+2bc+c^2$.

What do you get?

 

[lemonthree]

Just substitute $z=w+3i+2$ in $z^2-iw+5-2i=0$ and use the fact that $(a+b+c)^2=a^2+2ab+2ac+b^2+2bc+c^2$.

What do you get?

I got w = -2 + 2i or -2 - 7i, which is not the right answer

 

[skeeter]

Solve the simultaneous equations:
z = w + 3i + 2 and z² - iw + 5 - 2i = 0
giving z and w in the form (x + yi) where x and y are real.

$z = w + 3i + 2 \implies w = z-3i-2$

$z^2 - i(z-3i-2) + 5-2i = 0$

$z^2 - iz + 2 = 0$

now use the quadratic formula to solve for $z$ ... you should get

$z = 2i \implies w = -i-2$

or

$z =-i \implies w = -4i-2$