Question

solve for a and b in terms of c and d:
\[a+bi=\sqrt{c+di}\]

Answer 1

Square both sides and see what you have ... it's pretty straight-forward.

Answer 2

\[a^2-b^2+2abi=c+di\]
\[a^2-b^2=c\]
\[2ab=d\]
\[b=\frac{d}{2a}\]
\[a^2-\frac{d^2}{4a^2}=c\] is this right?

Answer 3

Yes, you're getting there. But remember you're looking for expressions for a and b

Answer 4

Now, notice, for instance in your last equation, if you multiply through by a^2, you'll have a quadratic in a^2, which you can solve.

Answer 5

I.e., a^4 - c.a^2 - d^2 = 0

Let X = a^2

then X^2 - cX - d = 0 hence X = .... and hence a = .....

Answer 6

\[4a^4-4ca^2-d^2=0\]
\[a^2=\frac{c\pm\sqrt{c^2+d^2}}{2}\]

Answer 7

Right. Now ... a, b, c and d are real numbers. So a^2 above must be non-negative. So only one solution works. Namely

a^2 = (1/2)(c + sqrt(c^2 + d^2))

The other solution would have a^2 being negative.

Answer 8

Now you can write down an expression for a and there will be two signs.

Then you substitute it back into your equations above to find an expression for b.

Answer 9

\[a=\pm\sqrt{\frac{c+\sqrt{c^2+d^2}}{2}}\]

Answer 10

ty

Answer 11

'good answer' appreciated.