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Question: If \[z = {i^9} + {i^{19}}\], then \[z\]is equal to A. \[0 + 0i\] B. \[1 + 0i\] C. \[0 + i\] ...

If z=i9+i19z = {i^9} + {i^{19}}, then zzis equal to
A. 0+0i0 + 0i
B. 1+0i1 + 0i
C. 0+i0 + i
D. 1+2i1 + 2i

Explanation

Solution

Here we will use the property of powers of complex numbers, that is in{i^n} equals to ii when nnis one more than the multiple of 4 and in{i^n} is equal to −i - i when nnis three more than the multiple of 4.

  • A complex number z=x+iyz = x + iy has real part xx and imaginary part yy.
  • Since i=−1i = \sqrt { - 1} therefore, i2=−1,i3=i×i2=i×(−1)=−i,i4=(i2)2=(−1)2=1{i^2} = - 1,{i^3} = i \times {i^2} = i \times ( - 1) = - i,{i^4} = {({i^2})^2} = {( - 1)^2} = 1
  • We break the power if ii is multiple of 44 because i4=1{i^4} = 1 and the rest powers can be allotted to another ii .
  • Also, in addition or subtraction of two complex numbers, always add the real part to the real part and the imaginary part to the imaginary part of the complex number.
  • Here we use the concept of the same base but different powers i.e. we have the same base ii therefore, powers can be separated, added or subtracted.

Complete step by step solution:
Given, z=i9+i19z = {i^9} + {i^{19}}
Since, 9=4×2+19 = 4 \times 2 + 1
i9{i^9}can be written as i4×2+1{i^{4 \times 2 + 1}}. Therefore from the property of powers of complex number when exponent of iiis one more than multiple of 44
i.e. i9=i4×2+1=(i2)4×(i)=1×i=i{i^9} = {i^{4 \times 2 + 1}} = {({i^2})^4} \times (i) = 1 \times i = i ...(1)...(1)
Now, 19=4×4+319 = 4 \times 4 + 3
Similarly i19{i^{19}} can be written as i4×4+3{i^{4 \times 4 + 3}}. Therefore from the property of powers of complex number when exponent of iiis three more than multiple of 4
i.e. i19=i4×4+3=(i4)4×(i3)=14×(−i)=−i{i^{19}} = {i^{4 \times 4 + 3}} = {({i^4})^4} \times ({i^3}) = {1^4} \times ( - i) = - i ...(2)...(2)
Substituting the values of i9{i^9} and i19{i^{19}} from equations (1)(1) and (2)(2) zz can be simplified toz=i9+i19=i+(−i)=i−i=0z = {i^9} + {i^{19}} = i + ( - i) = i - i = 0
Write the obtained value of zz in the form of a+bia + bi.

Therefore, z=0+0iz = 0 + 0i.

Therefore, option (A) is the correct answer.

Note:
In these types of questions where power of ii is involved, the exponent should be simplified in terms of multiples of 44, as i4n=1{i^{4n}} = 1. Also all the multiples of 44 can directly be written equal to 11. Whenever we get 00 as the answer always write the answer in form of a complex number that indicates both the real and the imaginary part as 00 i.e. z=0+0iz = 0 + 0i