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I have the following complex number $ z = (40-9i)·(a+7i) $ and I want to find the number $ a $ such that $ \left| z \right| = 1025 $

I tried to solve it through a couple of methods such as developing $ z $ to represent the product, try to isolate $ a $, try to put the modulus formula in an equation-like fashion but I'm stuck in the same place.

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$$|z|=|(40-9i)(a+7i)|=|40-9i||a+7i|=41|a+7i|=1025$$ $$\therefore |a+7i|=25 \implies a=\pm 24$$

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  • $\begingroup$ In "polar form" we can write the two numbers as $a_1e^{i\theta}_1$ and $a_2e^{\theta_2}$ where $a_1$ and $a_2$ are the moduli of the two complex numbers and $\theta_1$ and $\theta_2$ are the "arguments" so that the product is $(a_1a_2)e^{i(\theta_1+ \theta_2)}$. That clearly has modulus $a_1a_2$. That is why Peter Foreman can just multiply the two moduli. $\endgroup$
    – user247327
    Mar 10 '19 at 16:03
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Well: $$(40-9i)(a+7i)=(40a+63)+(280-9a)i$$ So: $$1025^2=(40a+63)^2+(280-9a)^2$$ This is just a quadratic.

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  • $\begingroup$ What's the problem with this answer? $\endgroup$ Mar 10 '19 at 15:57

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