Question

Find the roots of Z^(2) + (2-3I)Z -6i

Answer 1

Use \[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{2a }\]
where \[a =1\] \[b =2-3i\]
\[c =-6i\]

Answer 2

U ll get \[z =\frac{ -(2-3i)\pm \sqrt{-5+12i} }{ 2 }\]

Answer 3

It is probably not enough to just fill in the abc-formula.
Mostly it is required to give the answer in a+bi form.
Now sqrt(-5+12i) = 2+3i, so\[z_1=(-2+3i+2+3i)/2 =3i\]and\[z_2=(-2+3i-2-3i)/2=-4/2=-2\]

Answer 4

@ZeHanz how did you go from sqrt(-5+12i) to 2+3i

Answer 5

@ZeHanz Can you answer my question please i am really confused =D

Answer 6

Sorry about that!
I admit I use my calculator sometimes for these square roots...but when it gives you a "nice" number, it is worth while trying to get it without the help of a calculator.

Here is how you could do that: we look for a number a+bi, such that (a+bi)²=-5+12i.
After expanding the brackets, you get a²-b²+2abi=-5+12i, so:

a²-b²=-5 and 2ab=12, so ab=6.
ab=6 doesn't leave may options open for "nice" numbers, so why not try a=2 and b=3?

Yes! they do the trick! 2²-3²=4-9=-5.

I wouldn't know another way of doing it, besides halving the argument and taking the square root of the magnitude. In this case that doesnt work well, because the argument of -5+12i is itself not a "nice" angle.

I hope your a little less confused now!