MTH632 Quiz by Saman
- Type
- Quiz
- Course
- MTH632 Complex Analysis and Differential Geometry
- Size
- PDF 160 KB
- Updated
- 23 Oct 2026
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Q1. Express the complex number
)i in exponential form.
To express the complex number )i in exponential form, we first find the polar form of
)i and then square it.
- Find the modulus (r):
3) r z
- Find the argument :
We use the fact that the complex number is in the second quadrant (since ( 3) is negative and
( )i is positive). The argument ( ) is given by:
arg( ) tan z
Thus, we can write:
2 cos sin z i i
- Square the complex number:
Now square (z):
2 cos sin z i
Using the formula for powers of complex numbers in polar form: cos 2 sin 2 z i
4 cos sin z i
Thus, the exponential form of z is:
i
z e
Q2. Solution:
To express ( ( ) Im( ) Re( ) f x z z z z zz in the form f z u x y iv x y where
z x iy and z x iy let’s analyze each term and simplify.
- Write ( z ) and ( z in terms of ( x ) and ( y ):
z x iy
z x iy
The imaginary part of ( ), (Im z z y
The real part of ( ), (Re z z x
- Substitute and simplify each term:
First term: (
Im( ) z z
Im( ) z z x iy y xy iy
Second term: (
Re( ) z z
Re( ) z z x iy x x ixy
Third term: (
zz
zz x iy x iy x y
- Combine all terms to form ( f(z) ):
Now, we add up the three expressions we obtained:
Im( ) Re( ) f z z z z z zz f z xy iy x ixy x y
Group real and imaginary parts:
Real part ( u(x, y) ):
u x y xy x x y x xy y
Imaginary part ( v(x, y) ):
v x y y xy xy y
Q3. Find 4th roots of -1.
Solution:
Step 1: Express (-1) in Polar Form
The complex number (-1) has a modulus of (1) (since (|-1| = 1)) and an argument of ( ) (since it
lies on the negative real axis).
So, we can write:
1 1 cos( ) sin( ) i or in exponential form,
ie
Step 2: Apply De Moivre's Theorem for the 4th Roots
The fourth roots of a complex number ( i z re are given by:
for 0,1,2,3 k
i
k w r e k
In our case, (r = 1) and ( , so the 4th roots are:
for 0,1,2,3 k k i i
k w e e k
Step 3: Calculate Each Root
Now, we calculate each root by substituting (k = 0, 1, 2, 3):
- For (k = 0):
cos sin i
w e i i
- For (k = 1):
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