Pls. help me solve this complex number question, sir. Find the square root of the complex number (z^2 - (1 - i)z + 7i - 4 = 0. thanks sir. Also sir, its the square roots of the complex, not the roots, I solved it this way: using quadratic formula: z = (−b ± √(b²−4ac)) / (2a)
where: a = 1, b = −(1−i), c = (7i−4)
z = ((1−i) ± √((i−1)²−4(1)(7i−4))) / 2
z = ((1−i) ± √((i²−2i+1)−(28i−16))) / 2
z = ((1−i) ± √(−2i−28i+16)) / 2
z = ((1−i) ± √(16−30i) / 2
z = ((1−i) ± √(25−30i+9i²) / 2
z = ((1−i) ± √((5−3i)²) / 2
z = ((1−i) ± (5−3i)) / 2
z = 3 − 2i
z = −2 + i
but the lecturer said its wrong, he said that those are the roots and not the square root of the equation. so pls help sir, Thanks
You need the square roots of 3-2i and -2+i.
I will find the roots of 3-2i, then leave the roots of -2+i to you.
(a+bi)² = 3-2i
a² - b² + 2abi = 3-2i
a² - b² = 3
2ab = -2
b = -1/a
a² - 1/a² = 3
a^4 - 1 = 3a²
a^4 - 3a²- 1 = 0
This is a quadratic in a². Using the quadratic formula , we find
a² = (3+√13)/2 , where we reject (3-√13)/2 , because a must be real .
Thus , a =(±) √(3+√13)/√2
using b = -1/a , we get
b = (∓) √(3-√13)/√2 , where the sign of b is chosen to be opposite to that of a .
This gives the two square roots of 3-2i
I don't see any way to further simplify the expressions for a and b.
The two square roots of -2+i are found similarly.