Question:
A complex number \(z\) among the following which does not satisfy
\[
z^3+27i=0
\]
is:
A complex number \(z\) among the following which does not satisfy
\[
z^3+27i=0
\]
is:
Show Hint
For equations of the form \(z^n=w\), convert \(w\) into polar form and then use the \(n\)-th root formula to find all possible roots.
Updated On: Jun 18, 2026
- \(\dfrac{3\sqrt{3}-3i}{2}\)
- \(-3i\)
- \(\dfrac{3\sqrt{3}+3i}{2}\)
- \(\dfrac{-3\sqrt{3}+3i}{2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Rewrite the equation.
Given, \[ z^3+27i=0 \] So, \[ z^3=-27i. \] Now, \[ -27i=27\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right). \]
Step 2: Find the cube roots.
The cube roots are \[ z=3\left[\cos\left(\frac{\frac{3\pi}{2}+2k\pi}{3}\right) +i\sin\left(\frac{\frac{3\pi}{2}+2k\pi}{3}\right)\right], \] where \[ k=0,1,2. \]
Step 3: Calculate all roots.
For \(k=0\), \[ z=3\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) \] \[ z=3i. \] For \(k=1\), \[ z=3\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right) \] \[ z=3\left(-\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \] \[ z=\frac{-3\sqrt{3}-3i}{2}. \] For \(k=2\), \[ z=3\left(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\right) \] \[ z=3\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \] \[ z=\frac{3\sqrt{3}-3i}{2}. \]
Step 4: Compare with the options.
The roots obtained are \[ 3i,\quad \frac{-3\sqrt{3}-3i}{2},\quad \frac{3\sqrt{3}-3i}{2}. \] Among the given options, \[ \frac{3\sqrt{3}+3i}{2} \] does not satisfy the equation.
However, according to the marked answer in the question image, option (1) is selected.
Step 5: Final conclusion.
Mathematically, the value which does not satisfy the equation is \[ \boxed{\frac{3\sqrt{3}+3i}{2}} \]
Given, \[ z^3+27i=0 \] So, \[ z^3=-27i. \] Now, \[ -27i=27\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right). \]
Step 2: Find the cube roots.
The cube roots are \[ z=3\left[\cos\left(\frac{\frac{3\pi}{2}+2k\pi}{3}\right) +i\sin\left(\frac{\frac{3\pi}{2}+2k\pi}{3}\right)\right], \] where \[ k=0,1,2. \]
Step 3: Calculate all roots.
For \(k=0\), \[ z=3\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) \] \[ z=3i. \] For \(k=1\), \[ z=3\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right) \] \[ z=3\left(-\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \] \[ z=\frac{-3\sqrt{3}-3i}{2}. \] For \(k=2\), \[ z=3\left(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\right) \] \[ z=3\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \] \[ z=\frac{3\sqrt{3}-3i}{2}. \]
Step 4: Compare with the options.
The roots obtained are \[ 3i,\quad \frac{-3\sqrt{3}-3i}{2},\quad \frac{3\sqrt{3}-3i}{2}. \] Among the given options, \[ \frac{3\sqrt{3}+3i}{2} \] does not satisfy the equation.
However, according to the marked answer in the question image, option (1) is selected.
Step 5: Final conclusion.
Mathematically, the value which does not satisfy the equation is \[ \boxed{\frac{3\sqrt{3}+3i}{2}} \]
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