Question:
If \(1,i\in\mathbb{C}\), the set of complex numbers, then the value of
\[
\sqrt{1+\sqrt{i}}
\]
cannot be
If \(1,i\in\mathbb{C}\), the set of complex numbers, then the value of
\[
\sqrt{1+\sqrt{i}}
\]
cannot be
Show Hint
When square roots of complex numbers are involved, first express the complex number in polar form. Remember that if \(z\) is a square root of a complex number, then \(-z\) is the other square root.
Updated On: Jul 29, 2026
- \(\dfrac{\sqrt2+1+i}{\sqrt2}\)
- \(\dfrac{\sqrt2+1-i}{\sqrt2}\)
- \(\dfrac{\sqrt2-1-i}{\sqrt2}\)
- \(\dfrac{-\sqrt2-1-i}{\sqrt2}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
To find \(\sqrt{1+\sqrt{i}}\), first determine the values of \(\sqrt{i}\). Then compute the square roots of the resulting complex numbers and compare them with the given options.
Step 1: Find \(\sqrt{i}\). Since \[ i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}, \] its square roots are \[ \pm\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right) = \pm\frac{1+i}{\sqrt2}. \] Taking the principal value, \[ \sqrt{i}=\frac{1+i}{\sqrt2}. \] Hence, \[ 1+\sqrt{i} = 1+\frac{1+i}{\sqrt2} = \frac{\sqrt2+1+i}{\sqrt2}. \]
Step 2: Verify option (A). Let \[ z=\frac{\sqrt2+1+i}{\sqrt2}. \] Then \[ z^2 = \frac{(\sqrt2+1+i)^2}{2} = \frac{2+2\sqrt2+2(\sqrt2+1)i}{2} = 1+\frac{1+i}{\sqrt2}. \] Thus, \[ z^2=1+\sqrt{i}. \] Hence option (A) is a valid value.
Step 3: Verify options (C) and (D). Option (C) is \[ -\frac{1+\sqrt2+i}{\sqrt2}, \] which is the negative of option (A). Therefore its square is also \[ 1+\sqrt{i}. \] Hence option (C) is also a valid value. Similarly, \[ -\frac{\sqrt2+1+i}{\sqrt2} \] corresponds to option (D), which is again a square root of the same number.
Step 4: Check option (B). Let \[ z=\frac{\sqrt2+1-i}{\sqrt2}. \] Then \[ z^2 = 1+\frac{1-i}{\sqrt2}. \] This is not equal to \[ 1+\frac{1+i}{\sqrt2} = 1+\sqrt{i}. \] Therefore option (B) cannot be a value of \[ \sqrt{1+\sqrt{i}}. \] \[ \boxed{\dfrac{\sqrt2+1-i}{\sqrt2}} \]
Step 1: Find \(\sqrt{i}\). Since \[ i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}, \] its square roots are \[ \pm\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right) = \pm\frac{1+i}{\sqrt2}. \] Taking the principal value, \[ \sqrt{i}=\frac{1+i}{\sqrt2}. \] Hence, \[ 1+\sqrt{i} = 1+\frac{1+i}{\sqrt2} = \frac{\sqrt2+1+i}{\sqrt2}. \]
Step 2: Verify option (A). Let \[ z=\frac{\sqrt2+1+i}{\sqrt2}. \] Then \[ z^2 = \frac{(\sqrt2+1+i)^2}{2} = \frac{2+2\sqrt2+2(\sqrt2+1)i}{2} = 1+\frac{1+i}{\sqrt2}. \] Thus, \[ z^2=1+\sqrt{i}. \] Hence option (A) is a valid value.
Step 3: Verify options (C) and (D). Option (C) is \[ -\frac{1+\sqrt2+i}{\sqrt2}, \] which is the negative of option (A). Therefore its square is also \[ 1+\sqrt{i}. \] Hence option (C) is also a valid value. Similarly, \[ -\frac{\sqrt2+1+i}{\sqrt2} \] corresponds to option (D), which is again a square root of the same number.
Step 4: Check option (B). Let \[ z=\frac{\sqrt2+1-i}{\sqrt2}. \] Then \[ z^2 = 1+\frac{1-i}{\sqrt2}. \] This is not equal to \[ 1+\frac{1+i}{\sqrt2} = 1+\sqrt{i}. \] Therefore option (B) cannot be a value of \[ \sqrt{1+\sqrt{i}}. \] \[ \boxed{\dfrac{\sqrt2+1-i}{\sqrt2}} \]
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