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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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]^

Vedclass · Accepted Answer

$-122$

Vedclass · Answer

(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 $3b-2a=$

Similar Questions

If $a+ib = \frac{(x+i)^{2}}{2x^{2}+1}$,prove that $a^{2}+b^{2} = \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$.

If ${z_1} = 1 - i$ and ${z_2} = - 2 + 4i$,then $\text{Im} \left( \frac{z_1 z_2}{z_1} \right) = $

Evaluate: $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$.

The solution of the equation $|z| - z = 1 + 2i$ is

If $n$ is a positive integer,then which of the following relations is false?

Vedclass Test Series

Exam Paper Generator

Online Exam Module