Author Topic: Complex Number Question (Read 1249 times) Tweet Share

running_tape · « **on:** March 27, 2011, 01:20:48 pm »

I'm having problems finding the square roots of 1 + i using cartesian methods. I have two equations with 'a' and 'b', I've substituted one into the other and found 'a'. But when I substituted 'a' into the equation to find 'b', my answer for 'b' did not match the answer in the book (the Essentials, if you're wondering). Only when I solve for 'b' in the same manner as 'a' did I get the answer.

Sorry if my explanation of the situation is complicated. Maybe it's better if I see someone else's working. Help will be well appreciated.

luken93 · « **Reply #1 on:** March 27, 2011, 01:30:30 pm »

Quote from: running_tape on March 27, 2011, 01:20:48 pm

I'm having problems finding the square roots of 1 + i using cartesian methods. I have two equations with 'a' and 'b', I've substituted one into the other and found 'a'. But when I substituted 'a' into the equation to find 'b', my answer for 'b' did not match the answer in the book (the Essentials, if you're wondering). Only when I solve for 'b' in the same manner as 'a' did I get the answer.

Sorry if my explanation of the situation is complicated. Maybe it's better if I see someone else's working. Help will be well appreciated.

Post up the question?

Mao · « **Reply #2 on:** March 27, 2011, 03:05:52 pm »

Let the roots of 1+i be a+bi

$\begin{align*} (a+b\imath)^2 & = 1 + \imath \\ (a^2-b^2) + 2ab\imath & = 1 + \imath\end{align*}$
equating real and imaginary parts
$\begin{align*} a^2 - b^2 & = 1 \qquad \cdots (1) \\ 2ab & = 1 \qquad \cdots (2) \\ a & = \frac{1}{2b} \qquad \cdots \mbox{from }(2) \\ \frac{1}{4b^2} - b^2 & = 1 \qquad \cdots \mbox{substitute into }(1) \\ b^4 + b^2 - \frac{1}{4} & = 0 \\ b^2 & = \frac{-1 + \sqrt{2}}{2} \qquad \cdots \mbox{discarding negative, since }b^2>0 \\ b & = \pm \sqrt{\frac{-1+\sqrt{2}}{2}} \\ a & = \pm \frac{1}{\sqrt{-2+2\sqrt{2}}} = \pm \sqrt{\frac{1+\sqrt{2}}{2}}\\ \end{align*}$

enpassant · « **Reply #3 on:** March 27, 2011, 03:31:03 pm »

Quote from: Mao on March 27, 2011, 03:05:52 pm

Let the roots of 1+i be a+bi

$\begin{align*} (a+b\imath)^2 & = 1 + \imath \\ (a^2-b^2) + 2ab\imath & = 1 + \imath\end{align*}$
equating real and imaginary parts
$\begin{align*} a^2 - b^2 & = 1 \qquad \cdots (1) \\ 2ab & = 1 \qquad \cdots (2) \\ a & = \frac{1}{2b} \qquad \cdots \mbox{from }(2) \\ \frac{1}{4b^2} - b^2 & = 1 \qquad \cdots \mbox{substitute into }(1) \\ b^4 - b^2 - \frac{1}{4} & = 0 \\ b^2 & = \frac{1 + \sqrt{2}}{2} \qquad \cdots \mbox{discarding negative, since }b^2>0 \\ b & = \frac{\sqrt{1+\sqrt{2}}}{2} \\ a & = \frac{1}{\sqrt{1+\sqrt{2}}} \\ \end{align*}$

$b & = \sqrt{\frac{1+\sqrt{2}}{2}}$

should be another root

Mao · « **Reply #4 on:** March 27, 2011, 03:41:56 pm »

Quote from: enpassant on March 27, 2011, 03:31:03 pm

Quote from: Mao on March 27, 2011, 03:05:52 pm
Let the roots of 1+i be a+bi

$\begin{align*} (a+b\imath)^2 & = 1 + \imath \\ (a^2-b^2) + 2ab\imath & = 1 + \imath\end{align*}$
equating real and imaginary parts
$\begin{align*} a^2 - b^2 & = 1 \qquad \cdots (1) \\ 2ab & = 1 \qquad \cdots (2) \\ a & = \frac{1}{2b} \qquad \cdots \mbox{from }(2) \\ \frac{1}{4b^2} - b^2 & = 1 \qquad \cdots \mbox{substitute into }(1) \\ b^4 - b^2 - \frac{1}{4} & = 0 \\ b^2 & = \frac{1 + \sqrt{2}}{2} \qquad \cdots \mbox{discarding negative, since }b^2>0 \\ b & = \frac{\sqrt{1+\sqrt{2}}}{2} \\ a & = \frac{1}{\sqrt{1+\sqrt{2}}} \\ \end{align*}$

$b & = \sqrt{\frac{1+\sqrt{2}}{2}}$

should be another root

quite right, fixed now.

running_tape · « **Reply #5 on:** March 27, 2011, 07:33:14 pm »

Thanks, guys.

That was what I did. But the answer says:

Quote

(plus minus sign) $(\sqrt\frac{1+\sqrt{2}}{2}} + \sqrt{\frac{\sqrt2-1} {2}} \imath)$

Is it possible that it's a typo?

Mao · « **Reply #6 on:** March 27, 2011, 09:08:06 pm »

Quote from: running_tape on March 27, 2011, 07:33:14 pm

Thanks, guys.

That was what I did. But the answer says:
Quote
(plus minus sign) $(\sqrt\frac{1+\sqrt{2}}{2}} + \sqrt{\frac{\sqrt2-1} {2}} \imath)$

Is it possible that it's a typo?

No, the book's right. Turns out I've made another mistake in there. It's fixed now.

running_tape · « **Reply #7 on:** March 27, 2011, 09:57:46 pm »

Understood now. Thanks for all your help!

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