Solve the equation: z^2+iz+10=6z
I have no idea where to even begin...
I have no idea where to even begin...
-
Just rewrite it slightly and apply the quadratic formula
z^2 -(6 - i)z + 10 = 0
z = [(6 - i) +/- sqrt((6 - i)^2 - 40)]/2
z = 4 - 2i or z = 2 + i
~~~~~~~~~~~~~~
That last step can be found without a calculator if needed
(6 - i)^2 - 40 = 35 - 12i - 40 = -5 - 12i
This in 3rd quadrant with angle interpreted as -pi < arg < pi
It follows that the positive square root of this will have arg in 4th quadrant
Here is my homemade shortcut for square roots (when in integer coordinate form)
If z^2 = a + bi and r = sqrt(a^2 + b^2) as usual, then
sqrt[(r + a)/2] + sqrt[(r - a)/2] This gives the numbers in the square root.
But if z^2 in 3rd or 4th, must consider arg to for correct signs
To find positive sqrt(-5 - 12i), use r = 13, a = -5
sqrt[(13 + -5)/2] + sqrt[(13 - -5)/2] yields 2, 3 but expect fourth quadrant,
so result is 2 - 3i. Returning to your example,
z = [(6 - i) +/- (2 - 3i)]/2 = 4 - 2i or z = 2 + i
Regards - Ian
z^2 -(6 - i)z + 10 = 0
z = [(6 - i) +/- sqrt((6 - i)^2 - 40)]/2
z = 4 - 2i or z = 2 + i
~~~~~~~~~~~~~~
That last step can be found without a calculator if needed
(6 - i)^2 - 40 = 35 - 12i - 40 = -5 - 12i
This in 3rd quadrant with angle interpreted as -pi < arg < pi
It follows that the positive square root of this will have arg in 4th quadrant
Here is my homemade shortcut for square roots (when in integer coordinate form)
If z^2 = a + bi and r = sqrt(a^2 + b^2) as usual, then
sqrt[(r + a)/2] + sqrt[(r - a)/2] This gives the numbers in the square root.
But if z^2 in 3rd or 4th, must consider arg to for correct signs
To find positive sqrt(-5 - 12i), use r = 13, a = -5
sqrt[(13 + -5)/2] + sqrt[(13 - -5)/2] yields 2, 3 but expect fourth quadrant,
so result is 2 - 3i. Returning to your example,
z = [(6 - i) +/- (2 - 3i)]/2 = 4 - 2i or z = 2 + i
Regards - Ian