$$z= \sqrt{x^2+y^2}+ i\frac{\sqrt{2xy}}{x}-i+2i\sqrt{xy}$$ $(x,y>0)$
How do I solve this and find $|z|$
How do I solve equation and find $|z|$
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$\begingroup$
complex-numbers
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0Hint: $|a/b|=|a|/|b|$. – 2017-02-01
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0What equation? That looks like a definition of $z$. Hint: write it as $z=a+ib$ with $a,b \in \mathbb{R}$ then $|z|^2=a^2+b^2\,$. – 2017-02-01
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0If $z=x+iy$ this has no solution. Typos? – 2017-02-04
1 Answers
1
If $\space\exists\space\text{s}\in\mathbb{C}$:
$$\left|\text{s}\right|=\left|\Re\left(\text{s}\right)+\Im\left(\text{s}\right)i\right|=\sqrt{\Re^2\left(\text{s}\right)+\Im^2\left(\text{s}\right)}\tag1$$
So, in your example ( assuming that $x\space\wedge\space y\in\mathbb{R}$ ):
$$\left|\text{z}\right|=\left|\Re\left(\text{z}\right)+\Im\left(\text{z}\right)i\right|=\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag2$$
And we know that:
- $$\Re\left(\text{z}\right)=\sqrt{x^2+y^2}\tag3$$
- $$\Im\left(\text{z}\right)=\frac{\sqrt{2xy}}{x}-1+2\sqrt{xy}\tag4$$
So, we get:
$$\left|\text{z}\right|=\sqrt{\left(\sqrt{x^2+y^2}\right)^2+\left(\frac{\sqrt{2xy}}{x}-1+2\sqrt{xy}\right)^2}\tag5$$