Question

-8-6i, find the square root​

Answer 1

Answer:

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Answer 2

Given ,

Complex number (z) = -8 - 6i

Real part $\sf(Re_{(z)})$ = -8

Imaginary part $\sf (Im_{(z)})$ = - 6 < 0

Modules of complex number is given by :

$\pink{ \large \sf \fbox{ \fbox{ |z| = \sqrt{ {(a)}^{2} + {(b)}^{2} }} }}$

$\sf |z| = \sqrt{ {( - 8)}^{2} + {( - 6)}^{2} } \\ \\ \sf= \sqrt{64 + 36} \\ \\ \sf = \sqrt{100} \\ \\ \sf= 10$

Now , We know that , the square root of z if the imaginary part is less than 0 is given by :

$\sf \red{ \fbox{ \fbox{ \sqrt{z} = ± \bigg( \sqrt{ \frac{ |z| + Re_{(z)} }{2} } - i \sqrt{ \frac{ |z| - Re_{(z)}}{2} } \: \bigg) \: \: }}}$

$\sf = ± \bigg( \sqrt{ \frac{ 10 +( - 8) }{2} } - i \sqrt{ \frac{10 - ( - 8)}{2} } \: \bigg) \\ \\ \sf = ±( \sqrt{ \frac{2}{2} } - i \sqrt{ \frac{18}{2} } \: ) \\ \\ \sf = ±( \sqrt{1} - i \sqrt{9} ) \\ \\ \sf = ±(1 - 3i)$

Hence , the square root of given complex number is $\large \fbox{ \fbox{ ±(1 - 3i) }}$