If (a+ib)^{frac{1}{4}}=2+3i,then 3b-2a= | vedclass

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(B) Given $(a+ib)^{\frac{1}{4}}=2+3i$. Raising both sides to the power of $4$,we get: $(a+ib)=(2+3i)^4$. Expanding the right side: $(a+ib)=[(2+3i)^2]^

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$-122$

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(B) Given $(a+ib)^{\frac{1}{4}}=2+3i$. Raising both sides to the power of $4$,we get: $(a+ib)=(2+3i)^4$. Expanding the right side: $(a+ib)=[(2+3i)^2]^2 = [4+9i^2+12i]^2$. Since $i^2=-1$,we have: $(a+ib)=[4-9+12i]^2 = [-5+12i]^2$. Expanding further: $a+ib = (-5)^2 + (12i)^2 + 2(-5)(12i) = 25 - 144 - 120i = -119 - 120i$. Comparing the real and imaginary parts,we get $a=-119$ and $b=-120$. Now,calculating $3b-2a$: $3b-2a = 3(-120) - 2(-119) = -360 + 238 = -122$.

If $(a+ib)^{\frac{1}{4}}=2+3i$,then $3b-2a=$

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