find the square root of complex number 8-15i
Answers
Answer:
(3i- 5)/√2 or (5-3i)/√2
Step-by-step explanation:
ley say square root of complex number 8-15i is
a + ib
then squaring it
a^2 + i^2b^2 + 2abi
a^2 - b^2 + 2abi
comparing with
8 - 15i
a^2-b^2 = 8
2ab = - 15
b = -15/2a
a^2 - (-15/2a)^2 = 8
4a^4 - 225 = 32a^2
4a^4 -32a^2 - 225 = 0
4a^4 - 50a^2 + 18a^2 - 225 = 0
2a^2(2a^2 - 25)+(2a^2 - 25) = 0
(2a^2 + 9) (2a^2 - 25) = 0
a^2 = -9/2. or a^2 = 25/2
a = 3i/√2. a = 5/√2
b = -5i/√2. b = -3/√2
a + ib = (5 - 3i)/√2
or
a + ib = (3i - 5)/√2
square root of complex number 8-15i is
(3i- 5)/√2 or (5-3i)/√2
Answer:
5-3i/root2
Step-by-step explanation:
a^2 - (-15/2a)^2 = 8
4a^4 - 225 = 32a^2
4a^4 -32a^2 - 225 = 0
4a^4 - 50a^2 + 18a^2 - 225 = 0
2a^2(2a^2 - 25)+(2a^2 - 25) = 0
(2a^2 + 9) (2a^2 - 25) = 0
a^2 = -9/2. or a^2 = 25/2
a = 3i/√2. a = 5/√2
b = -5i/√2. b = -3/√2