Question

-15-8i squre root of alzebra

Answer 1

Assume −15−8i−−−−−−−√=(a+ib)−15−8i=(a+ib).

⇒(a+ib)2=−15−8i⇒(a+ib)2=−15−8i

⇒a2+i2b2+2iab=−15−8i⇒a2+i2b2+2iab=−15−8i

⇒a2−b2+2iab=−15−8i⇒a2−b2+2iab=−15−8i

⇒ab=−4;a2−b2=−15⇒ab=−4;a2−b2=−15

Now, ⇒a2−(−4a)2=−15⇒a2−(−4a)2=−15

⇒a2−16a2=−15⇒a2−16a2=−15

⇒a4+15a2−16=0⇒a4+15a2−16=0

⇒a=±4,a=1⇒a=±4,a=1

Now, b=−4a=−4±4=±1b=−4a=−4±4=±1

Therefore, −15−8i−−−−−−−√=±(4−i)


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