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z^2=5-12i
Where z=a+bi
How to do it, help ls
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z = a + bi ---> = = = (since = = = // <--- since = -1)
Now you have, from the condition, an equality of these two complex numbers:
= 5 - 12i.
It means that the real parts are equal and the imaginary parts are equal:
= 5, (1)
2ab = -12. (2)
It is the system of two equations for the two unknowns "a" and "b".
To solve it, express b = = from (2) and substitute it into (1). You will get
= 5, or
- = 5.
Multiply both sides by a^2. You will get
= 0.
Factor the left side
= 0.
The equation deploys in two independent equations
1. = 0 ---> = 9 ---> a = +/- 3 ---> b = -6/a = -/+ 2.
2. = 0 ---> = -4 ---> No real solutions.
Answer. The solutions of the original equation = 5 - 12i are these two complex numbers: = 3 - 2i or/and = -3 + 2i.
In other words, the square root of the complex number 5 - 12i in the complex domain are these two complex numbers: 3 - 2i and -3 + 2i.
There is a bunch of my lessons on complex numbers
- Complex numbers and arithmetical operations on them
- Complex plane
- Addition and subtraction of complex numbers in complex plane
- Multiplication and division of complex numbers in complex plane
- Raising a complex number to an integer power
- How to take a root of a complex number
- Solution of the quadratic equation with real coefficients on complex domain
- How to take a square root of a complex number
- Solution of the quadratic equation with complex coefficients on complex domain
in this site.