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I need to take the absolute value of the following expression:

$$c_1e^{ic_2}+ic_3e^{-ic_6}\sqrt{c_4+c_5i}$$

One strategy might be to separate it into the form $a+ib$, but how to handle the term inside the square root?

(all the constants $c_n$ are real)

1 Answers1

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Well, the square root of a complex number is:

$$\sqrt{\text{c}_4+\text{c}_5\cdot i}=\Re\left(\sqrt{\text{c}_4+\text{c}_5\cdot i}\right)+\Im\left(\sqrt{\text{c}_4+\text{c}_5\cdot i}\right)\cdot i\tag1$$

Where:

  • $$\Re\left(\sqrt{\text{c}_4+\text{c}_5\cdot i}\right)=\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}+\text{c}_4}{2}}\tag2$$
  • $$\Im\left(\sqrt{\text{c}_4+\text{c}_5\cdot i}\right)=\pm\space\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}-\text{c}_4}{2}}\tag3$$

So, we get:

  • $$\Re\left(\text{c}_1\cdot\exp\left(\text{c}_2\cdot i\right)+\text{c}_3\cdot\exp\left(-\text{c}_6\cdot i\right)\cdot\sqrt{\text{c}_4+\text{c}_5\cdot i}\cdot i\right)=$$ $$\Re\left(\text{c}_1\cdot\exp\left(\text{c}_2\cdot i\right)\right)+\Re\left(\text{c}_3\cdot\exp\left(-\text{c}_6\cdot i\right)\cdot\sqrt{\text{c}_4+\text{c}_5\cdot i}\cdot i\right)=$$ $$\text{c}_1\cdot\cos\left(\text{c}_2\right)+\text{c}_3\cdot\left(\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}+\text{c}_4}{2}}\cdot\sin\left(\text{c}_6\right)\mp\space\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}-\text{c}_4}{2}}\cdot\cos\left(\text{c}_6\right)\right)\tag4$$
  • $$\Im\left(\text{c}_1\cdot\exp\left(\text{c}_2\cdot i\right)+\text{c}_3\cdot\exp\left(-\text{c}_6\cdot i\right)\cdot\sqrt{\text{c}_4+\text{c}_5\cdot i}\cdot i\right)=$$ $$\Im\left(\text{c}_1\cdot\exp\left(\text{c}_2\cdot i\right)\right)+\Im\left(\text{c}_3\cdot\exp\left(-\text{c}_6\cdot i\right)\cdot\sqrt{\text{c}_4+\text{c}_5\cdot i}\cdot i\right)=$$ $$\text{c}_1\cdot\sin\left(\text{c}_2\right)+\text{c}_3\cdot\left(\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}+\text{c}_4}{2}}\cdot\cos\left(\text{c}_6\right)\pm\space\sqrt{\frac{\sqrt{\text{c}_4^2+\text{c}_5^2}-\text{c}_4}{2}}\cdot\sin\left(\text{c}_6\right)\right)\tag5$$
Jan Eerland
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