I have to solve :$$(i+z)^{16}-(7+7i)(i+z)^8+25i=0$$ I consider this equation like a quadratic equation and find its delta and solutions. But i can't solve two equations :$$(i+z)^8=4+3i$$ $$(i+z)^8=3+4i$$ I really need some helps to solve completely this equation. Thanks.
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3Convert the RHS to polar form, find the eighth root(s), and solve for $z$. – Joshua Wang Feb 11 '21 at 03:07
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If you don't want to use polar coordinates, the eighth roots can be found by repeated square root.
We get the square root $$ (x+iy)^2 = a+ib $$ when $$ x = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} + \sqrt{\sqrt{a^2+b^2} - b} \right) $$
$$ y = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} - \sqrt{\sqrt{a^2+b^2} - b} \right) $$
if the real parts $a,b$ are positive, so are the $x,y$ above are also positive. Switching $a,b$ to $b,a$ gives a different result.
Will Jagy
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Don't you think it could be easier to write $a+bi=\frac 1{\sqrt{a^2+b^2}}(\cos(x)+i\sin(x))$ and identify $x$ ? – Claude Leibovici Feb 11 '21 at 03:26
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@ClaudeLeibovici sure. Still, it all can be done with lots of square root signs applying to real quantities. If there is no interest from the OP, little is lost. – Will Jagy Feb 11 '21 at 03:31