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In complex number we have real and imaginary part mixed and \(\sqrt{-1}\) is the basic unit and denoted by \(i\). In the given question we have to find value of k for which the equation will be valid.

How many integers \(k\) are there for which \((1-i)^k=2^k\) ?

(A) One

(B) None

(C) Two

(D) More than one.

Source

Competency

Difficulty

Suggested Book

ISI entrance B. Stat. (Hons.) 2003 problem 5

Complex numbers

6 out of 10

Challenges and thrills of pre-college mathematics

First hint

The complex number \((1-i)\) can be rationalized by multiplying numerator and denominator by \(1+i\).

And we will get

\((1-i)=(1-i)\frac{(1+i)}{1+i}=\frac{2}{1+i}\)

Second Hint

Now we will have \((\frac{2}{1+i})^k=2^k\)

so, \((1+i)^k=1\), this is only possible when k=0,

So \(k\) can have only one value, The option (A) is correct.

- https://www.cheenta.com/problem-based-on-divisibility-cmi-2015-problem-3/
- https://www.youtube.com/watch?v=P4ZYA4XCQoM

Content

[hide]

In complex number we have real and imaginary part mixed and \(\sqrt{-1}\) is the basic unit and denoted by \(i\). In the given question we have to find value of k for which the equation will be valid.

How many integers \(k\) are there for which \((1-i)^k=2^k\) ?

(A) One

(B) None

(C) Two

(D) More than one.

Source

Competency

Difficulty

Suggested Book

ISI entrance B. Stat. (Hons.) 2003 problem 5

Complex numbers

6 out of 10

Challenges and thrills of pre-college mathematics

First hint

The complex number \((1-i)\) can be rationalized by multiplying numerator and denominator by \(1+i\).

And we will get

\((1-i)=(1-i)\frac{(1+i)}{1+i}=\frac{2}{1+i}\)

Second Hint

Now we will have \((\frac{2}{1+i})^k=2^k\)

so, \((1+i)^k=1\), this is only possible when k=0,

So \(k\) can have only one value, The option (A) is correct.

- https://www.cheenta.com/problem-based-on-divisibility-cmi-2015-problem-3/
- https://www.youtube.com/watch?v=P4ZYA4XCQoM

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