Why is $1+i^3$ equal to $1-i$?

Because $i^3 = -i$, so $1+i^3 = 1-i$. It is just a simple substitution based on the cyclic property of $i$.

Can I use polar form to solve this?

Yes, you absolutely can! $1+i = \sqrt{2}e^{i\pi/4}$. Then $(1+i)^n = 2^{n/2}e^{in\pi/4}$. This is a valid approach, but the conjugate property is much faster and more elegant for this specific problem.

Is there a trick to the options? Why is the answer $n_1 > 0, n_2 > 0$?

The options are designed to make you look for a specific relationship between $n_1$ and $n_2$. However, the expression is real for *all* positive integers. So, the condition is simply that they must be positive integers.

JEE Advanced 1996

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: For positive integers $n_{1}, n_{2}$ , the value of the expression $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$ , where $i = - 1$ is a real number if and only if

Select Answer:

Visualized Solution

The Given Expression

Given expression: $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$
We need to find the condition on $n_{1}, n_{2}$ for this to be purely real.

Powers of $i$

Recall the fundamental property: $i = - 1$
$i^{2} = - 1$

Simplifying $i^{3}$

$i^{3} = i^{2} \cdot i$
$i^{3} = (- 1) \cdot i = - i$

Simplifying $i^{5}$

$i^{4} = (i^{2})^{2} = (- 1)^{2} = 1$
$i^{5} = i^{4} \cdot i = 1 \cdot i = i$

Simplifying $i^{7}$

$i^{7} = i^{4} \cdot i^{3}$
$i^{7} = 1 \cdot (- i) = - i$

Substituting Simplified Powers

Substitute the values back:
$(1 + i)^{n_{1}} + (1 - i)^{n_{1}} + (1 + i)^{n_{2}} + (1 - i)^{n_{2}}$

Visualizing $1 + i$

Let $z = 1 + i$
This is a point in the first quadrant of the complex plane.

Visualizing $1 - i$

The term $1 - i$ is the complex conjugate of $z$ , denoted as $\overset{z}{ˉ}$ .
It is a reflection of $z$ across the real axis.

Sum of Conjugates

For any complex number $z$ , $z + \overset{z}{ˉ} = 2 Re (z)$ .
The imaginary parts cancel out perfectly.

Powers of Conjugates

Property: $(\overset{z}{ˉ})^{n} = \overline{(z^{n})}$
Therefore, $z^{n} + (\overset{z}{ˉ})^{n} = z^{n} + \overline{(z^{n})} = 2 Re (z^{n})$ .

Analyzing the First Part

Apply the property to the first two terms:
$(1 + i)^{n_{1}} + (1 - i)^{n_{1}} = 2 Re ((1 + i)^{n_{1}})$
This is purely real for any integer $n_{1}$ .

Analyzing the Second Part

Apply the property to the next two terms:
$(1 + i)^{n_{2}} + (1 - i)^{n_{2}} = 2 Re ((1 + i)^{n_{2}})$
This is also purely real for any integer $n_{2}$ .

The Final Conclusion

Total expression = Real Number + Real Number = Real Number.
This holds true for all positive integers $n_{1}$ and $n_{2}$ .
Condition: $n_{1} > 0, n_{2} > 0$ .

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The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

The given expression is:

(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}

To simplify this, we must first address the powers of the imaginary unit

i

, where

i = - 1

. Recall that the powers of

i

follow a cyclic pattern of length four:

i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1

Simplifying the Terms

Using the cyclic property, we evaluate the individual components of the expression:

1 + i^{3} = 1 - i

1 + i^{5} = 1 + i

1 + i^{7} = 1 - i

Substituting these back into the original expression, we obtain:

(1 + i)^{n_{1}} + (1 - i)^{n_{1}} + (1 + i)^{n_{2}} + (1 - i)^{n_{2}}

Applying Symmetry

Observe that the expression is now composed of pairs of the form

z^{n} + \overset{z}{ˉ}^{n}

, where

z = 1 + i

and

\overset{z}{ˉ} = 1 - i

. These are complex conjugates of each other.

For any complex number

z

, the sum

z + \overset{z}{ˉ} = 2 Re (z)

, which is always a real number. This property extends to higher powers:

z^{n} + \overset{z}{ˉ}^{n} = 2 Re (z^{n})

Conclusion

Since

(1 + i)^{n_{1}} + (1 - i)^{n_{1}}

is real and

(1 + i)^{n_{2}} + (1 - i)^{n_{2}}

is real, their sum must also be real.

Therefore, the entire expression is real for any positive integers $n_{1}$ and $n_{2}$ . The complexity of the initial expression is resolved entirely through the symmetry of complex conjugates.

Animated Solution for Mathematics - Complex Numbers: For positive integers $n_{1}, n_{2}$ , the value of the expression $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$ , where $i = - 1$ is a real number if and only if

Select Answer:

Visualized Solution

The Given Expression

Powers of $i$

Simplifying $i^{3}$

Simplifying $i^{5}$

Simplifying $i^{7}$

Substituting Simplified Powers

Visualizing $1 + i$

Visualizing $1 - i$

Sum of Conjugates

Powers of Conjugates

Analyzing the First Part

Analyzing the Second Part

The Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Simplifying the Terms

Applying Symmetry

Conclusion

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Visualized Solution

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Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i3

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i5

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i7

Substituting Simplified Powers

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+i

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1−i

Sum of Conjugates

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The Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

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