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\[\sqrt{9-9i}\]
solve z^2 = 9-9i
@jayzdd How do you mean?
you can use the trig form of the complex number z^2 = sqrt(9^2 + 9^2) ( cos(-45) + i sin(-45) )
you must be knowing that angles add up when you multiply two complex numbers
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\[(r_1\angle \theta_1 )(r_2\angle \theta_2) = r_1r_2 \angle (\theta_1+\theta_2)\]
Finding \(\sqrt{9-9i}\) is same as finding a number which when multiplied by itself gives you \(9-9i\)
that is, finding some \(z\) such that \[z*z = 9-9i\]
|dw:1445578809479:dw|
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