SOLUTION: How do i solve the equation _ What is the square root of 4-7i?

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Question 397698: How do i solve the equation _ What is the square root of 4-7i?
Found 3 solutions by Alan3354, robertb, richard1234:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
That's not an equation, there's no equal sign.
Use DeMoivre's Theorem to find the sq root.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Suggestion:
Let sqrt%284-7i%29++=a+%2B+bi, so that 4-7i+=+%28a%2Bbi%29%5E2+=+a%5E2+-+b%5E2+%2B+2abi.
==> a%5E2+-+b%5E2+=+4+, and 2ab+=+-7. After substitution,
a%5E2+-+49%2F%284a%5E2%29+=+4, or 4a%5E4-16a%5E2+-+49+=+0.
==> a%5E2+=+%2816+%2B-+sqrt%281040+%29%29%2F8+=+%284+%2B-+sqrt%2865%29%29%2F2+. Eliminate the negative value to get
a%5E2+=+%284+%2B+sqrt%2865%29%29%2F2.
Get the square root of this, and plug into 2ab+=+-7.
Good luck!

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
It is indeed possible to solve it using Euler's formula, by multiplying/dividing by sqrt%2865%29 to obtain a unit vector and obtaining:

where cos+%28theta%29+=+4%2Fsqrt%2865%29 and sin+%28theta%29+=+-7%2Fsqrt%2865%29.

However, the easiest way is probably to assume that the square root of 4+-+7i is a+%2B+bi for real numbers a,b. Squaring both sides,

%28a%2Bbi%29%5E2+=+4+-+7i
a%5E2+%2B2abi+-+b%5E2+=+4+-+7i

Equating the real and imaginary parts, we get the system of equations

a%5E2+-+b%5E2+=+4
2ab+=+-7 --> ab = -7/2.

Using the substitution b = -7/2a,

a%5E2+-+%28-7%2F2a%29%5E2+=+4
a%5E2+-+49%2F4a%5E2+-+4+=+0
4a%5E4+-+16a%5E2+-+49+=+0

Use the quadratic formula to solve for a%5E2 then take the square root. Note that a%5E2 must be positive based on our assumption that a and b are real numbers. After that, the rest is just arithmetic.