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If I had an expression $|a+bi+\sqrt(c+di)|^{2}$, how would I evaluate this? I know how to evaluate the magnitude of a complex expression or the root of a complex expression separately but not when they are combined.

bob
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  • Calculate $\sqrt{c+di}$ according to your definition of the square root for complex numbers (principal value, maybe?), then the problem reduces to the $u+iv$ case you know. – dxiv Sep 19 '17 at 04:57
  • Thanks for the help. I was able to figure out the solution. Just for reference, I used this post as reference: https://math.stackexchange.com/questions/664962/what-is-the-square-root-of-complex-number-i. – bob Sep 19 '17 at 05:11
  • Glad it helped. That said, I hope you also noted the unfortunately, this cannot be answered definitively caveat in the accepted answer to the linked question, and the reference to the principal value of the complex square root in my previous comment. Your question is not complete unless you spell out what the definition is for the $\sqrt{,\cdot,}$ complex square root being used. – dxiv Sep 19 '17 at 05:23

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Well, when we have $\text{z}_1\space\wedge\space\text{z}_2\in\mathbb{C}$:

$$\text{D}:=\left|\text{z}_1+\sqrt{\text{z}_2}\right|\le\left|\text{z}_1\right|+\left|\sqrt{\text{z}_2}\right|=\left|\text{z}_1\right|+\sqrt{\left|\text{z}_2\right|}=$$ $$\sqrt{\Re^2\left(\text{z}_1\right)+\Im^2\left(\text{z}_1\right)}+\sqrt{\sqrt{\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)}}=$$ $$\sqrt{\Re^2\left(\text{z}_1\right)+\Im^2\left(\text{z}_1\right)}+\left(\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)\right)^\frac{1}{4}\tag1$$

And the square root of a complex number:

$$\sqrt{\text{z}_2}=\Re\left(\sqrt{\text{z}_2}\right)+\Im\left(\sqrt{\text{z}_2}\right)\cdot i\tag2$$

Where:

  • $$\Re\left(\sqrt{\text{z}_2}\right)=\sqrt{\frac{\sqrt{\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)}+\Re\left(\text{z}_2\right)}{2}}\tag3$$
  • $$\Im\left(\sqrt{\text{z}_2}\right)=\pm\space\sqrt{\frac{\sqrt{\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)}-\Re\left(\text{z}_2\right)}{2}}\tag4$$

So, when we want to find:

$$\text{D}=\sqrt{\Re^2\left(\text{z}_1+\sqrt{\text{z}_2}\right)+\Im^2\left(\text{z}_1+\sqrt{\text{z}_2}\right)}\tag5$$

Where:

  • $$\Re\left(\text{z}_1+\sqrt{\text{z}_2}\right)=\Re\left(\text{z}_1\right)+\Re\left(\sqrt{\text{z}_2}\right)=$$ $$\Re\left(\text{z}_1\right)+\sqrt{\frac{\sqrt{\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)}+\Re\left(\text{z}_2\right)}{2}}\tag6$$
  • $$\Im\left(\text{z}_1+\sqrt{\text{z}_2}\right)=\Im\left(\text{z}_1\right)+\Im\left(\sqrt{\text{z}_2}\right)=$$ $$\Im\left(\text{z}_1\right)\pm\space\sqrt{\frac{\sqrt{\Re^2\left(\text{z}_2\right)+\Im^2\left(\text{z}_2\right)}-\Re\left(\text{z}_2\right)}{2}}\tag7$$
Jan Eerland
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