Question

Find the square root of 12 + 20 iota.
[Please Solve step by Step]​

Answer 1

Answer:

Let −21−20i=x+iy

⟹−21−20i=(x+iy)2=x2−y2+2ixy

Equating real and imaginary parts we get

x2−y2=−21 ------(1)

2xy=−20 ------(2)

Therefore, (x2+y2)2=(x2−y2)2+4x2y2=(−21)2+(−20)2=841

⟹x2+y2=841=29 ------(3)

adding (1) and (3) we get

2x2=8

⟹x=±4=±2

Substituting x in (2) we get

y=∓5

Therefore, the square root of −21−20i is ±(2−5i)

Step-by-step explanation:

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